Category: Calculus

The Wallis Product

Problem 1. By considering the integral $\displaystyle I(n) := \int_0^{\pi} \sin^n(x)\, \mathrm dx$ , prove the Wallis product

$\displaystyle \frac 21 \cdot \frac 23 \cdot \frac 43 \cdot \frac 45 \cdot \cdots := \lim_{n \to \infty} \prod_{k=1}^n \frac{2n \cdot 2n}{(2n-1)(2n+1)} = \frac{\pi}{2}.$

(Click for Solution)

Solution. We leave it as an exercise to check that $I(0) = \pi$ and $I(1) = 2$ . These observations motivate an expression of $I(n+2)$ in terms of $I(n)$ . To that end, we integrate by parts to obtain

$\begin{aligned} & \int \sin^{n+2}(x)\, \mathrm dx = \int \sin(x) \sin^{n+1}(x)\, \mathrm dx \\ &= \underbrace{-{\cos(x)}}_{\mathrm I} \underbrace{\sin^{n+1}(x)}_{\mathrm S} - \int \underbrace{-{\cos(x)}}_{\mathrm I} \underbrace{(n+1) \sin^n(x) \cos(x) }_{\mathrm D}\, \mathrm dx \\ &= -{\cos(x)} \sin^{n+1}(x) + (n+1) \int \cos^2(x) \sin^n(x)\, \mathrm dx \\ &= -{\cos(x)} \sin^{n+1}(x) + (n+1) \int (1-\sin^2(x)) \sin^n(x)\, \mathrm dx \\ &= -{\cos(x)} \sin^{n+1}(x) + (n+1) \left( \int \sin^n(x)\, \mathrm dx - \int \sin^{n+2}(x)\, \mathrm dx\right). \end{aligned}$

By algebruh,

$\displaystyle (n+2) \int \sin^{n+2}(x)\, \mathrm dx = -{\cos(x)} \sin^{n+1}(x) + (n+1) \int \sin^n(x)\, \mathrm dx.$

Plugging in the limits of $[0, \pi]$ ,

$\displaystyle (n+2) \int_0^{\pi} \sin^{n+2}(x)\, \mathrm dx = (n+1) \int_0^{\pi} \sin^n(x)\, \mathrm dx.$

Hence,

$\displaystyle \frac{I(n+2)}{I(n)} = \frac{n+1}{n+2}.$

In particular,

$\displaystyle I(2k+2) = \frac{2k+1}{2k+2} \cdot I(2k) \quad\Rightarrow \quad I(2n+2) = \pi \cdot \prod_{k=1}^n \frac{2k-1}{2k}$

Similarly,

$\displaystyle I(2n+1) = 2 \cdot \prod_{k=1}^n \frac{2k}{2k+1}.$

Dividing the terms appropriately,

$\displaystyle \prod_{k=1}^n \frac{2k \cdot 2k}{(2k-1) (2k+1)} = \frac{\pi}{2} \cdot \frac{I(2n+1)}{I(2n+2)}.$

It suffices to prove that $I(2n+1)/I(2n+2) \to 1$ . Since $0 \leq \sin(x) \leq 1$ for $x \in [0, \pi]$ ,

$0 \leq \sin^{n+1}(x) \leq \sin^n(x) \leq \sin^{n-1}(x) \leq 1.$

By the monotonicity of integration, $I(n+1) \leq I(n)$ so that

$\displaystyle 1 \leq \frac{I(2n+1)}{I(2n+2)} \leq \frac{I(2n)}{I(2n+2)} = \frac{2n+2}{2n+1} \to 1.$

By the squeeze theorem, $I(2n+1)/I(2n+2) \to 1$ , as required.

—Joel Kindiak, 21 Apr 25, 2129H

April 21, 2025
The Intermediate Value Derivative

Problem 1. Let $f : [a, b] \to \mathbb R$ differentiable. Assume $f'(a) < f'(b)$ . Prove that for any $m \in (f'(a), f'(b))$ , there exists $c \in [a, b]$ such that $f'(c) = m$ .

(Click for Solution)

Solution. We observe that $f'(c) = m \iff (f(x)-mx)'(c) = 0$ . Define the differentiable function $g : [a, b] \to \mathbb R$ by $g(x) = f(x) - mx$ . Then it suffices to find $c \in [a, b]$ such that $g'(c) = 0$ .

Since $g$ is differentiable on $[a, b]$ , it is continuous on $[a, b]$ . By the extreme value theorem, there exists $c \in [a, b]$ such that $g(x) \geq g(c)$ for any $x \in [a, b]$ . We claim that $g'(c) = 0$ .

Firstly, we note that $g'(a) = f'(a) - m < 0$ , so that there exists $\delta_1 > 0$ such that $g(c) \leq g(a+\delta_1) < g(a)$ . Similarly, $g'(b) = f'(b) - m > 0$ , so that there exists $\delta_2 > 0$ such that $g(c) \leq g(b-\delta_2) < g(b)$ . Hence, $g(c) \neq g(a), g(b)$ . Therefore, $c \in (a, b)$ , which implies $g'(c) = 0$ , as required.

Remark 1. Any function that satisfies the conclusion of the intermediate value theorem is called a Darboux function. The intermediate value theorem demonstrates that continuous functions are Darboux functions. This problem demonstrates that derivatives are automatically Darboux functions, even if they are not continuous.

—Joel Kindiak, 22 Jan 25, 1234H

April 3, 2025
Several Laborious Integrals

Problem 1. Evaluate the integral $\displaystyle \int \frac{(2x+1) \ln(9x+5)}{(3x+2)^4} \, \mathrm dx$ .

(Click for Solution)

Solution. We first evaluate $\displaystyle \int \frac{(2x+1) }{(3x+2)^4} \, \mathrm dx$ for simplicity. Integrating by parts,

$\displaystyle \begin{aligned} \int \frac{(2x+1) }{(3x+2)^4} \, \mathrm dx &= \underbrace{ \frac{(3x+2)^{-3}}{-3 \cdot 3} }_{\mathrm I} \cdot \underbrace{(2x+1)}_{\mathrm S} - \int \underbrace{ \frac{(3x+2)^{-3}}{-3 \cdot 3} }_{\mathrm I} \cdot \underbrace{2}_{\mathrm D}\, \mathrm dx \\ &= -\frac{2x+1}{9(3x+2)^3} + \frac 29 \int (3x+2)^{-3}\, \mathrm dx \\ &= -\frac{2x+1}{9(3x+2)^3} + \frac 29 \cdot \frac{(3x+2)^{-2}}{-2 \cdot 3} +C \\ &= -\frac{2x+1}{9(3x+2)^3} - \frac 1{27(3x+2)^2} +C \\ &= -\frac{3(2x+1)}{27(3x+2)^3} - \frac {3x+2}{27(3x+2)^3} +C \\ &= -\frac{3(2x+1) + 3x+2}{27(3x+2)^3} +C \\ &= -\frac{9x+5}{27(3x+2)^3} + C.\end{aligned}$

Therefore, integrating by parts again,

$\displaystyle \begin{aligned} & \int \frac{(2x+1) \ln(9x+5)}{(3x+2)^4} \, \mathrm dx \\ &= \underbrace{-\frac{9x+5}{27(3x+2)^3}}_{\mathrm I} \cdot \underbrace{\ln(9x+5)}_{\mathrm S} - \int \underbrace{-\frac{9x+5}{27(3x+2)^3}}_{\mathrm I}\cdot \underbrace{\frac 1{9x+5} \cdot 9}_{\mathrm D}\, \mathrm dx \\ &= -\frac{(9x+5)\ln(9x+5)}{27(3x+2)^3} + \frac 13 \int \frac{1}{(3x+2)^3}\, \mathrm dx \\ &= -\frac{(9x+5)\ln(9x+5)}{27(3x+2)^3} + \frac 13 \int (3x+2)^{-3}\, \mathrm dx \\ &= -\frac{(9x+5)\ln(9x+5)}{27(3x+2)^3} + \frac 13 \cdot \frac 13 \cdot \frac{(3x+2)^{-2}}{-2} + C \\ &= -\frac{(9x+5)\ln(9x+5)}{27(3x+2)^3} - \frac 1{18 (3x+2)^2} + C.\end{aligned}$

Problem 2. Evaluate the integral $\displaystyle \int_4^5 \frac{9x^3 - 14x^2 + 8x - 11}{(x-1)(x-2)(x^2+3)} \, \mathrm dx$ .

(Click for Solution)

Solution. We decompose into a sum of partial fractions as follows:

$\displaystyle \frac{9x^3 - 14x^2 + 8x - 11}{(x-1)(x-2)(x^2+3)} = \frac{A}{x-1} + \frac{B}{x-2} + \frac{Cx+D}{x^2+3}.$

Assuming we have evaluated $A,B,C,D$ , integrating yields

$\displaystyle \begin{aligned} &\int_4^5 \frac{9x^3 - 14x^2 + 8x - 11}{(x-1)(x-2)(x^2+3)} \, \mathrm dx \\ &= \int_4^5 \left(A \cdot \frac{1}{x-1} + B \cdot \frac{1}{x-2} + \frac{C}{2} \cdot \frac{2x}{x^2+3} + D \cdot \frac{1}{(\sqrt 3)^2 + x^2}\right)\, \mathrm dx \\ &= \left[A \ln|x-1| + B \ln |x-2| + \frac C2 \ln(x^2+3) + \frac{D}{\sqrt 3} \tan^{-1}\left( \frac{x}{\sqrt 3}\right)\right]_4^5 \\ &= A \ln \frac 43 + B \ln \frac 32 + \frac C2 \ln \frac{28}{19} + \frac{D}{\sqrt 3} \left( \tan^{-1}\frac{5}{\sqrt 3} - \tan^{-1}\frac{4}{\sqrt 3}\right). \end{aligned}$

It remains to evaluate $A,B,C,D$ . Cross-multiplying yields

$\displaystyle \begin{aligned} 9x^3 - 14x^2 + 8x - 11 &= A(x-2)(x^2+3) \\ &\phantom{=} + B(x-1)(x^2+3)\\ &\phantom{=} + (Cx+D)(x-1)(x-2).\end{aligned}$

Setting $x = 1$ yields $A = 2$ .
Setting $x = 2$ yields $B = 3$ .

Comparing the terms involving $x^3$ ,

$9 = A + B + C \quad \Rightarrow \quad C = 4.$

Comparing the constant terms,

$-11 = -6A + (-3B) + 2D \quad \Rightarrow \quad D = 5.$

Therefore, the definite integral evaluates to

$\begin{aligned} &\int_4^5 \frac{9x^3 - 14x^2 + 8x - 11}{(x-1)(x-2)(x^2+3)} \, \mathrm dx \\ &=2 \ln \frac 43 + 3 \ln \frac 32 + \frac 42 \ln \frac{28}{19} + \frac{5}{\sqrt 3} \left( \tan^{-1}\frac{5}{\sqrt 3} - \tan^{-1}\frac{4}{\sqrt 3}\right)\\ &= 2 \ln \frac {112}{57} + 3 \ln \frac 32 + \frac{5}{\sqrt 3} \left( \tan^{-1}\frac{5}{\sqrt 3} - \tan^{-1}\frac{4}{\sqrt 3}\right) \\ &= 2 \ln \frac {112}{57} + 3 \ln \frac 32 + \frac{5}{\sqrt 3}\tan^{-1}\frac{\sqrt 3}{23},\end{aligned}$

where the last equality follows from the identity

$\displaystyle \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1} \left(\frac{a-b}{1+ab}\right).$

Problem 3. Given $a > 0$ and $b^2 - 4ac < 0$ , evaluate $\displaystyle \int \frac{1}{\sqrt{ax^2 + bx + c}}\, \mathrm dx$ . Deduce the value of

$\displaystyle \lim_{n \to \infty} \left( \frac{1}{\sqrt{n^2+1^2}} + \frac{1}{\sqrt{n^2+2^2}} + \cdots + \frac{1}{\sqrt{n^2+n^2}} \right).$

(Click for Solution)

Solution. For the integral, we first complete the square

$ax^2 + bx + c = a(x+h)^2 + k,$

where $h$ and $k$ are constants to be determined in terms of $a,b,c$ . To determine $h, k$ , we expand the right-hand side to obtain

$ax^2 + bx + c = ax^2 + (2ah)x + ah^2 + k.$

Comparing coefficients $h = b/(2a)$ and

$\begin{aligned} k = c - ah^2 &= c - a \left(\frac b{2a}\right)^2 \\ &= c - a \cdot \frac{b^2}{4a^2} \\ &= \frac{4ac- b^2}{4a} \\ &= \frac{-(b^2 - 4ac)}{4a} > 0. \end{aligned}$

Therefore,

$\begin{aligned} &\int \frac{1}{\sqrt{ax^2 + bx + c}}\, \mathrm dx \\ &= \int \frac{1}{\sqrt{a(x+h)^2 + k}}\, \mathrm dx \\ &= \frac{1}{\sqrt a} \int \frac{1}{\sqrt{(x+h)^2 + (\sqrt{k/a})^2 }}\, \mathrm dx \\ &= \frac 1{\sqrt a} \ln \left|(x+h) + \sqrt{(x+h)^2 + (\sqrt{k/a})^2} \right| + C \\ &= \frac 1{\sqrt a} \ln \left| x+\frac b{2a} + \sqrt{ \left(x+\frac b{2a} \right)^2 + \frac{4ac-b^2}{4a^2} } \right| + C. \end{aligned}$

For the limit, we factorise $n^2$ to obtain

$\displaystyle \begin{aligned} &\lim_{n \to \infty} \left( \frac{1}{\sqrt{n^2+1^2}} + \frac{1}{\sqrt{n^2+2^2}} + \cdots + \frac{1}{\sqrt{n^2+n^2}} \right) \\ &=\lim_{n \to \infty} \frac 1n \left( \frac{1}{\sqrt{1+\left( \frac 1n\right)^2}} + \frac{1}{\sqrt{1+\left( \frac 2n\right)^2}} + \cdots + \frac{1}{\sqrt{1+\left( \frac nn\right)^2}} \right) \\ &= \int_0^1 \frac{1}{\sqrt{1+x^2}}\, \mathrm dx, \end{aligned}$

where the last equality follows from the definition of integration. Setting $a = 1, b = 0, c = 1$ in the indefinite integral,

$\begin{aligned} \int_0^1 \frac{1}{\sqrt{1+x^2}}\, \mathrm dx &= \left[ \frac 1{\sqrt 1} \ln \left| x+ 0 + \sqrt{ (x+ 0 )^2 + \frac{4 \cdot 1 \cdot 1 - 0^2}{4 \cdot 1^2} } \right| \right]_0^1 \\ &= \left[ \ln \left| x + \sqrt{ x^2 + 1 } \right| \right]_0^1 \\ &= \ln ( 1 + \sqrt{ 2 } ) - 0 \\ &= \ln ( 1 + \sqrt{ 2 } ).\end{aligned}$

—Joel Kindiak, 24 Feb 25, 1800H

March 26, 2025

Calculus, math, mathematics, Physics, science
Cauchy’s True Utility

Let’s talk about infinite sums. Consider the following sums:

$\displaystyle \begin{aligned}s &= 1 + 1 + 1 + 1 + \cdots, \\ t &= 1 + 2 + 3 + 4 + \cdots, \\ u &= 1 +2 + 4 + 8 + \cdots, \\ v &= 1 + \frac 12 + \frac 14 + \frac 18 + \cdots. \end{aligned}$

Since we add infinitely many terms, our sums should be infinite, right? That is the case for $s, t, u$ , but surprisingly, not for $v$ . But let’s define what we mean by an infinite sum properly.

We will adopt the convention, as with other topics, that $\mathbb N = \mathbb N^+$ , so that $0 \notin \mathbb N$ . Let $\mathbb K$ be an ordered field.

Definition 1. Let $\{x_n\}$ be a $\mathbb K$ -sequence. Inductively define the partial sums sequence $\{s_n\}$ by

$s_1 := x_1,\quad s_{n+1} := s_n + x_{n+1}.$

Often, we denote

$\begin{aligned} s_n &= \sum_{k=1}^n x_k \equiv x_1 + x_2 + \cdots + x_n,\\ s_\infty &= \sum_{k=1}^\infty x_k = x_1+ x_2 + x_3 + \cdots. \end{aligned}$

We say that the series $s_\infty$ converges to a real number $s$ if $s_n \to s$ , and diverges otherwise. In this case, we write

$\displaystyle s_\infty = \lim_{n \to \infty} s_n = s.$

Example 1. Given $\{x_n\} = \{1\}$ with partial sums $s_n = 1 + \cdots + 1 = n$ , $s_n = n\to \infty$ , thus $s_\infty$ diverges.

Example 2. Given $\{x_n\} = \{r^{n-1}\}$ , the sequence of partial sums is given by

$s_n = \begin{cases}\frac{1-r^n}{1-r}, & \quad r \neq 1, \\ 1, & \quad r = 1. \end{cases}$

If $|r| < 1$ , then

$\displaystyle |r^n| = |r|^n \to 0 \quad \Rightarrow \quad s_\infty = \frac 1{1-r}.$

If $r > 1$ , then

$\displaystyle r^n \to \infty \quad \Rightarrow \quad s_n = \frac{ 1-r^n }{ 1-r } \to -\infty.$

If $r < -1$ , then $-r > 1$ implies

$\displaystyle \begin{aligned} s_{2k} &= \frac{1-r^{2k}}{1-r} = \frac{1-(-r)^{2k}}{1+(-r)} \to -\infty, \\ s_{2k+1} &= \frac{1-r^{2k+1}}{1-r} = \frac{1+(-r)^{2k+1}}{1+(-r)} \to \infty. \end{aligned}$

If $r = 1$ , then $s_n = n \to \infty$ . If $r = -1$ , then

$\displaystyle \begin{aligned} s_{2k} &= \frac{1-(-1)^{2k}}{1-(-1)} = \frac{1-1}{2} = 0 \to 0, \\ s_{2k+1} &= \frac{1-(-1)^{2k+1}}{1-(-1)} = \frac{1-(-1)}{2} = 1 \to 1. \end{aligned}$

However, if $s_\infty = s$ , then $0 = s = 1$ , a contradiction.

This gives us our first crucial convergence result.

Theorem 1. For any $r$ , the partial sums of $\{r^n\}$ , called the standard geometric series, converges to $1/(1-r)$ if and only if $|r| < 1$ . In summation notation,

$\displaystyle \sum_{k=1}^\infty r^k \equiv 1 + r + r^2 + r^3 + \cdots = \frac 1{1-r}$

if and only if $|r| < 1$ .

Example 3. In particular, setting $r = 1/2$ so that $|r| < 1$ ,

$\displaystyle 1 + \frac 12 + \frac 14 + \frac 18 + \cdots = \frac 1{1 - 1/2} = 2.$

But what other series converge? Recall that we write

$x_1 + x_2 + x_3 + \cdots = s$

if and only if the partial sums $s_n \to s$ . Let’s expand this latter idea in epsilontics.

Lemma 1. Suppose $\{s_n\}$ converges. Then for any $\epsilon > 0$ , there exists $N \in \mathbb N$ such that for $m,n > N$ ,

$|s_m - s_n| < \epsilon.$

Proof. Fix $\epsilon > 0$ . Then there exists $N \in \mathbb N$ such that

$\displaystyle m,n > N \quad \Rightarrow \quad |s_m - s|< \frac{\epsilon}{2}, \quad |s_n - s| < \frac{\epsilon}{2}.$

By the triangle inequality, for $m, n > N$ ,

$\displaystyle |s_m - s_n| \leq |s_m - s| + |s - s_n| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$

This is what we mean by a Cauchy sequence.

Definition 2. A $\mathbb K$ -sequence $\{x_n\}$ is Cauchy if for any $\epsilon > 0$ , there exists $N \in \mathbb N$ such that

$m,n > N \quad \Rightarrow \quad |x_m - x_n| < \epsilon.$

By Lemma 1, for any $\mathbb K$ -seqeunce $\{x_n\}$ , if $\{x_n\}$ converges, then $\{x_n\}$ is Cauchy. Does the reverse hold? Interestingly, when $\mathbb K = \mathbb R$ , the answer is Yes. That is one property that sets $\mathbb R$ apart from $\mathbb Q$ , and it boils down back to the completeness property.

Lemma 2. If $\{x_n\}$ is Cauchy, then $\{x_n\}$ is bounded.

Proof. Since $\{x_n\}$ is Cauchy, for $\epsilon = 1$ , there exists $N \in \mathbb N$ such that

$m,n > N \quad \Rightarrow \quad |x_m - x_n| < 1.$

Then for any $n > N$ ,

$|x_n| \leq |x_n - x_{N+1}| + |x_{N+1}| < 1 + |x_{N+1}|.$

This means that if we define $M := \max \{1,|x_1|, \cdots, |x_N|,|x_{N+1}|\}$ , then for any $n \in \mathbb N$ ,

$|x_n| \leq 1 + M.$

Thus, $\{x_n\}$ is bounded.

Theorem 2. Let $\{x_n\}$ be an $\mathbb R$ -sequence. Then $\{x_n\}$ converges if and only if $\{x_n\}$ is Cauchy.

Proof. We have proven $(\Rightarrow)$ and now turn our attention to prove $(\Leftarrow)$ .

Let $\{x_n\}$ be a Cauchy $\mathbb R$ -sequence. Fix $\epsilon > 0$ . What convergence theorems do we have to prove that $x_n$ converges?

We have the Bolzano-Weierstrass theorem, but that theorem asserts the convergence of a subsequence of $\{x_n\}$ , not the convergence of $\{x_n\}$ itself. Furthermore, this only works if $\{x_n\}$ is bounded. Thankfully, Lemma 2 tells us that this is, in fact, true. Therefore, there exists a subsequence $\{ x_{n_k} \}$ of $\{x_n\}$ such that $x_{n_k} \to x$ .

We somehow need to take advantage of the Cauchy property of $\{x_n\}$ . Fix $\epsilon > 0$ . Since $x_{n_k} \to x$ , there exists $K \in \mathbb N$ such that

$\displaystyle k > K \quad \Rightarrow \quad |x_{n_k} - x| < \frac{\epsilon}{2}.$

Since $\{x_n\}$ is Cauchy, there exists $N \in \mathbb N$ such that

$\displaystyle m,n > N \quad \Rightarrow \quad |x_m - x_n| < \frac{\epsilon}{2}.$

Define $M := \max\{K, N\}$ . Then for $n > n_{M+1}$ , since $n_{M+1} \geq M+1 > M$ ,

$\displaystyle |x_n - x| \leq |x_n - x_{n_{M+1}}| + |x_{n_{M+1}} - x| < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon.$

That’s our proof!

In terms of our main subject of study—infinite series—we get the Cauchy criterion for convergent series.

Corollary 1. Let $\{x_n\}$ be an $\mathbb R$ -sequence. Then $\displaystyle s_\infty \equiv \sum_{k=1}^\infty x_k$ converges if and only if $\{s_n\}$ is Cauchy.

We have established some convergent series, some divergent series, and a comprehensive test to check for convergence. But math people (and worse, engineers) are lazy and don’t have all the time to derive every detail from first principles. Are there some useful rules-of-thumb to help us create more convergent series?

That is the content, which is quite crucial in all honesty, of our next post.

—Joel Kindiak, 28 Dec 24, 1218H

March 19, 2025
What is the Exponential Unit?

You might have seen the number $e \approx 2.718$ in various mathematical discussions. What makes $e$ such a crucial number? It turns out to be the key number in calculus, because

$\displaystyle \frac{\mathrm d}{\mathrm dx} e^x = e^x.$

Our goal is to define $e$ rigorously, then prove this result, which is a highly nontrivial result.

Theorem 1. The sequence $\{e_n\}$ defined by $e_n := (1+1/n)^n$ converges. Thus, the exponential unit is defined by

$\displaystyle e := \lim_{n \to \infty} e^n \equiv \lim_{n \to \infty} \left(1 + \frac 1n\right)^n.$

Proof of Theorem 1. By the monotone convergence theorem, we need to show that $\{e_n\}$ is bounded above and non-decreasing, or bounded below and non-increasing.

We need to compute $e_{n+1}/e_n$ and show that this ratio satisfies either $\leq 1$ or $\geq 1$ . By definition,

$\displaystyle \begin{aligned} \frac{e_{n+1}}{e_n} = \frac{ (1+\frac{1}{n+1})^{n+1} }{(1+\frac{1}{n})^n} &= \frac{ (1+\frac{1}{n+1})^{n+1} }{(1+\frac{1}{n})^{n+1}} \cdot \left(1+\frac{1}{n} \right) \\ &= \left(\frac{ 1+\frac{1}{n+1}}{1+\frac{1}{n}}\right)^{n+1} \cdot \left(1+\frac{1}{n} \right) \\ &= \left( \frac{n(n+2)}{(n+1)^2} \right)^{n+1} \cdot \left(1+\frac{1}{n} \right) \\ &= \left( 1 - \frac{1}{(n+1)^2} \right)^{n+1} \cdot \left(1+\frac{1}{n} \right). \end{aligned}$

Heuristically, we would think that $1/(n+1)^2 \to 0$ more quickly than $1/(n+1) \to 0$ , so that the whole limit $\to 1^+$ . But how do we formulate this idea rigorously?

With a stroke of genius, we will need Bernoulli’s inequality, which states that for real numbers $x > -1$ and integers $r \geq 0$ ,

$(1+x)^r \geq 1 + rx.$

We will prove this result later. For now, since $n+1 \geq 0$ and $x = -1/(n+1)^2 > -1$ , we can apply Bernoulli’s inequality to obtain

$\displaystyle \begin{aligned} \frac{ e_{n+1} }{ e_n } &= \left( 1 - \frac{1}{(n+1)^2} \right)^{n+1} \cdot \left(1+\frac{1}{n} \right) \\ &\geq \left( 1 + (n+1) \cdot \left(-\frac{1}{(n+1)^2} \right) \right) \cdot \left(1+\frac{1}{n} \right) \\ &= \left( 1 - \frac{1}{n+1} \right) \cdot \left(1+\frac{1}{n} \right) = \frac{n}{n+1} \cdot \frac{n+1}{n} = 1, \end{aligned}$

which implies that $e_{n+1} \geq e_n$ , which establishes that $\{e_n\}$ . This hints that we need to establish that $\{e_n\}$ is bounded above. Experimentally, we notice that $e_n \leq 3$ , so we aim to prove this result rigorously, via induction.

Now for $n = 1$ , we get

$\displaystyle e_1 = \left( 1 + \frac 11 \right)^1 = 2 \leq 3.$

Now inductively, we first suppose that $e_k \leq 3$ , and aim to prove that $e_{k+1} \leq 3$ . To that end,

$\begin{aligned} e_{k+1} &= \left(1 + \frac 1{k+1} \right)^{k+1} \\ &= \left(1 + \frac 1{k+1} \right)^{k} \cdot \left(1 + \frac 1{k+1}\right) \\ &= \left(1 + \frac 1{k} \right)^{k} \cdot \frac{ \left(1 + \frac 1{k+1} \right)^{k} }{ \left(1 + \frac 1{k} \right)^{k} } \cdot \left(1 + \frac 1{k+1}\right) \\ &= \left(1 + \frac 1{k} \right)^{k} \cdot \left( 1 - \frac{1}{(k+1)^2} \right)^k \cdot \left(1 + \frac 1{k+1}\right) \\ &\leq 3 \cdot \left( 1 - \frac{1}{k+1} \right)^k \cdot \left(1 + \frac 1{k+1}\right)^k \\ &\leq 3 \cdot \left( 1 - \frac{1}{(k+1)^2} \right)^k \leq 3 \cdot (1-0)^k = 3, \end{aligned}$

as required. Thus, $\{e_n\}$ is bounded above and non-decreasing. By the monotone convergence theorem, $\{e_n\}$ converges, as required.

Now, this proof hinges on Bernoulli’s inequality, which we will now prove using induction.

Lemma 1 (Bernoulli’s Inequality). For real numbers $x > -1$ and integers $r \geq 0$ ,

$(1+x)^r \geq 1 + rx.$

Proof of Bernoulli’s Inequality. We will prove by induction on $r$ . The result for $r = 0$ is obvious. Suppose now that

$(1+x)^k \geq 1+kx.$

We need to prove that $(1+x)^{k+1} \geq 1 + (k + 1)x$ . Then

$\begin{aligned} (1+x)^{k+1} &= (1+x)^k \cdot (1+x) \\ &\geq (1+kx)\cdot (1+x) \\ &= 1 + (k+1)x + kx^2 \\ &\geq 1 + (k+1)x + 0 \\ &= 1 + (k+1)x.\end{aligned}$

Now we aim to prove the real result of interest.

Theorem 2. $\displaystyle \frac{\mathrm d}{\mathrm dx} e^x = e^x$ .

Proof of Theorem 2. Fix $x \in \mathbb R$ . By the definition of the derivative, we need to take $h \to 0$ of the difference quotient

$\displaystyle \frac{e^{x+h} - e^x}{h} = \frac{e^x e^h - e^x}{h} = e^x \cdot \frac{e^h - 1}{h}.$

If we can prove that $(e^h - 1)/h \to 1$ as $h \to 0^+$ , then we will obtain

$\displaystyle \begin{aligned} \lim_{h \to 0^-} \frac{e^h - 1}{h} &= \lim_{h \to 0^+} \frac{e^{-h} - 1}{-h} \\ &= \lim_{h \to 0^+} \frac{1 - e^{-h} }{h} \\ &= \lim_{h \to 0^+} \left( e^{-h} \cdot \frac{e^h - 1}{h} \right) \\ &= \lim_{h \to 0^+} e^{-h} \cdot \lim_{h \to 0^+} \frac{e^h - 1}{h} \\ &= 1 \cdot 1 = 1, \end{aligned}$

so that $(e^h - 1)/h \to 1$ as $h \to 0$ in either direction and the result follows. To that end, let’s use the limit definition of $e$ to our advantage. Our first observation is that by a useful re-indexing,

$\displaystyle \begin{aligned} e^h &= \left(\lim_{n \to \infty} \left( 1 + \frac 1n \right)^n \right)^h \\ &= \lim_{n \to \infty} \left( 1 + \frac 1n \right)^{nh} \\ &= \lim_{n \to \infty} \left( 1 + \frac h{nh} \right)^{nh} \\ &= \lim_{n \to \infty} \left( 1 + \frac hn \right)^n, \end{aligned}$

since $h > 0$ . The expression $(1+h/n)^n$ suggests the use of the binomial theorem:

$\displaystyle \begin{aligned} \left( 1 + \frac hn \right)^n &= \sum_{k=0}^n {n \choose k} \frac{h^k}{n^k} = 1 + h + \sum_{k=2}^n {n \choose k} \frac{h^k}{n^k} \end{aligned}$

With the help of algebra and the triangle inequality,

$\displaystyle \begin{aligned} \left| \frac{\left( 1 + h/n \right)^n - 1}{h} - 1 \right| &\leq h \cdot \sum_{k=0}^{n-2} {n \choose k+2} \frac{ h^k }{n^{k+2}}.\end{aligned}$

We will need to take $h \to 0$ , so we can assume $|h| \leq 1/2$ . Let’s analyse each term inside the sum (perhaps that is where real analysis gets its name from). We observe that

$\begin{aligned} {n \choose k+2} \frac{ h^k }{n^{k+2}} &= \frac{n(n-1)\cdot \cdots \cdot (n-k-1)}{n^{k+2} } \cdot \frac{h^k}{(k+2)!} \\ &= \frac{n}{n} \cdot \frac{n-1}{n} \cdot \cdots \cdot \frac{n-k-1}{n} \cdot \frac{h^k}{(k+2)!} \\ &\leq 1^n \cdot (1/2)^k = (1/2)^k. \end{aligned}$

Using the formula for a geometric series that we will prove in just a moment,

$\displaystyle \begin{aligned} \sum_{k=0}^{n-2} {n \choose k+2} \frac{ h^k }{n^{k+2}} &\leq \sum_{k=0}^{n-2} (1/2)^k = \frac{1 - (1/2)^{n-1}}{1-1/2} \leq \frac{1 - 0}{1-1/2} = 2. \end{aligned}$

Therefore,

$\displaystyle \begin{aligned} \left| \frac{\left( 1 + h/n \right)^n - 1}{h} - 1 \right| &\leq h \cdot \sum_{k=0}^{n-2} {n \choose k+2} \frac{ h^k }{n^{k+2}} \leq 2 h.\end{aligned}$

Taking $n \to \infty$ ,

$\displaystyle \left| \frac{e^h - 1}{h} - 1 \right| = \lim_{n \to \infty} \left| \frac{\left( 1 + h/n \right)^n - 1}{h} - 1 \right| \leq \lim_{n \to \infty} (2h) = 2 h.$

Taking $h \to 0^+$ , the squeeze theorem yields

$\displaystyle \lim_{h \to 0^+} \left| \frac{e^h - 1}{h} - 1 \right| = 0.$

Therefore,

$\displaystyle \lim_{h \to 0^+} \frac{e^h - 1}{h} = 1,$

as required.

Lemma 2 (Geometric Series). For any $r \neq 0$ ,

$\displaystyle \sum_{k=0}^{n-1} r^k = \frac{1-r^n}{1-r}.$

Proof of Lemma 2. The proof is immediate by the observation

$\displaystyle \begin{aligned} (1-r) \sum_{k=0}^{n-1} r^k &= \sum_{k=0}^{n-1} r^k - \sum_{k=0}^{n-1} r^{k+1} \\ &= 1 + \sum_{k=1}^{n-1} r^k - \sum_{k=0}^{n-2} r^{k+1} - r^n \\ &= 1 + \sum_{k=1}^{n-1} r^k - \sum_{k=1}^{n-1} r^{k} - r^n \\ &= 1 - r^n.\end{aligned}$

Dividing yields the desired result.

We now have several simple real-analytic ingredients to explore infinite sums, which we shall do next time.

—Joel Kindiak, 26 Dec 24, 1845H

March 12, 2025
Calculus Recap (Polytechnic)

Limits

Problem 1. Evaluate the limit $\displaystyle \lim_{x \to 1} (x+1)$ .

(Click for Solution)

Solution. Substituting $x = 1$ ,

$\displaystyle \begin{aligned} \lim_{x \to 1} (x+1) &= 1 + 1 = 2. \end{aligned}$

Problem 2. Evaluate the limit $\displaystyle \lim_{x \to 1} \frac{x^2 - 1}{x-1}$ .

(Click for Solution)

Solution. Factorising for $x \neq 1$ ,

$\displaystyle \begin{aligned} \lim_{x \to 1} \frac{x^2 - 1}{x-1} &= \lim_{x \to 1} \frac{(x-1)(x+1)}{x-1} \\ &= \lim_{x \to 1} (x+1) = 2, \end{aligned}$

where the last $=$ follows from Problem 1.

Problem 3. Evaluate the limit $\displaystyle \lim_{x \to 1} \frac{\sqrt x - 1}{x-1}$ .

(Click for Solution)

Solution. Rationalising the numerator then substituting $x = 1$ ,

$\displaystyle \begin{aligned} \lim_{x \to 1} \frac{\sqrt x - 1}{x-1} &= \lim_{x \to 1} \frac{\sqrt x - 1}{x-1} \cdot \frac{\sqrt x + 1}{\sqrt x + 1} \\ &= \lim_{x \to 1} \frac{(\sqrt x)^2 - 1^2}{(x-1)(\sqrt x + 1)} \\ &= \lim_{x \to 1} \frac{x - 1}{(x-1)(\sqrt x + 1)} \\ &= \lim_{x \to 1} \frac 1{\sqrt x + 1} = \frac 1{\sqrt 1 + 1} = \frac 12.\end{aligned}$

Problem 4. Without using differentiation, compute the gradient of the tangent to $y =x^2$ at $(1, 1)$ .

(Click for Solution)

Solution. Denote $(x_1, y_1) = (1, 1)$ and $(x_2, y_2) = (x, x^2)$ . By definition of the gradient of the tangent,

$\displaystyle \begin{aligned} \lim_{x \to 1} \frac{\text{(rise)}}{\text{(run)}} &= \lim_{x \to 1} \frac{y_2 - y_1}{x_2 - x_1} = \lim_{x \to 1} \frac{x^2 - 1}{x - 1} = 2,\end{aligned}$

where the last $=$ follows from Problem 2.

Differentiation

Problem 5. Using differentiation, compute the gradient of the tangent to $y = x^2$ at $(1, 1)$ .

(Click for Solution)

Solution. Differentiating,

$\displaystyle \frac{\mathrm dy}{\mathrm dx} = \frac{\mathrm d}{\mathrm dx} (x^2) = 2x.$

Setting $x = 1$ , $\displaystyle \frac{\mathrm dy}{\mathrm dx} = 2 \cdot 1 = 2$ .

Problem 6. Defining $f(x) = x^2$ , evaluate $f''(1)$ .

(Click for Solution)

Solution. Differentiating twice,

$\begin{aligned} f(x) &= x^2, \\ f'(x) &= 2x, \\ f''(x) &= 2. \end{aligned}$

Setting $x = 1$ , $f''(1) = 2$ .

Problem 7. Evaluate $\displaystyle \frac{\mathrm d}{\mathrm dx}\left(\ln(x^2+9) \right)$ .

(Click for Solution)

Solution. Employing the chain rule,

$\displaystyle \begin{aligned} \frac{ \mathrm d }{ \mathrm dx }\left( \ln(x^2+1) \right) &= \frac 1{x^2 + 9} \cdot \frac{\mathrm d}{\mathrm dx}(x^2 + 1) \\ &= \frac 1{x^2 + 9} \cdot 2x \\ &= \frac { 2x }{x^2 + 9}. \end{aligned}$

Problem 8. Evaluate $\displaystyle \frac{\mathrm d}{\mathrm dx}\left(\sin \left(\ln(x^2+9) \right) \right)$ .

(Click for Solution)

Solution. Employing the chain rule and Problem 7,

$\displaystyle \begin{aligned} \frac{\mathrm d}{\mathrm dx}\left(\sin \left(\ln(x^2+1) \right) \right) &= \cos\left(\ln(x^2+9) \right) \cdot \frac{\mathrm d}{\mathrm dx}\ln(x^2+9) \\ &= \cos\left(\ln(x^2+9) \right) \cdot \frac {2x}{x^2 + 9} \\ &= \frac {2x \cos\left(\ln(x^2+9) \right) }{x^2 + 9}. \end{aligned}$

Problem 9. Evaluate $\displaystyle \frac{\mathrm d}{\mathrm dx} \left( (\ln x) \cdot (x^2 + 9) \right)$ .

(Click for Solution)

Solution. Employing the product rule,

$\displaystyle \begin{aligned} \frac{\mathrm d}{\mathrm dx} \left( (\ln x) \cdot (x^2 + 9) \right) &= (x^2 + 9) \cdot \frac{\mathrm d}{\mathrm dx} ( \ln x ) + \ln x \cdot \frac{\mathrm d}{\mathrm dx} (x^2 + 9) \\ &= (x^2 + 9) \cdot \frac 1x + \ln x \cdot (2x) \\ &= x + \frac 9x + 2x \ln x. \end{aligned}$

Problem 10. Evaluate $\displaystyle \frac{\mathrm d}{\mathrm dx} \left( \frac{\ln x}{x^2 + 9}\right)$ .

(Click for Solution)

Solution. Employing the quotient rule,

$\displaystyle \begin{aligned} \frac{\mathrm d}{\mathrm dx} \left( \frac{\ln x}{x^2 + 9} \right) &= \frac{(x^2 +9) \cdot \frac{\mathrm d}{\mathrm dx} ( \ln x ) - \ln x \cdot \frac{\mathrm d}{\mathrm dx} (x^2 + 9) }{(x^2 + 9)^2} \\ &= \frac{ (x^2 + 9) \cdot \frac 1x - \ln x \cdot 2x}{(x^2 + 9)^2 } \\ &= \frac{x^2 + 9 -2x^2 \ln x}{x(x^2 + 9)^2}. \end{aligned}$

Integration

Problem 11. Evaluate $\displaystyle \int \left(2\ln x + \frac 5{x^2+9}\right)\, \mathrm dx$ .

(Click for Solution)

Solution. Applying the linearity of integration and using common formulas (here and here),

$\displaystyle \begin{aligned} \int \left(2\ln x + \frac 5{x^2+9}\right)\, \mathrm dx &= \int \left(2 \cdot\ln x + 5 \cdot \frac 1{x^2 + 9}\right)\, \mathrm dx \\ &= 2 \cdot \int \ln x \, \mathrm dx + 5 \cdot \int \frac 1{3^2 + x^2} \, \mathrm dx \\ &= 2 \cdot (x \ln x - x) + 5 \cdot \frac 1{3} \tan^{-1} \left(\frac x3 \right) + C \\ &= 2x \ln x - 2x + \frac 53 \tan^{-1} \left(\frac x3 \right) + C. \end{aligned}$

Problem 12. Evaluate $\displaystyle \int_1^2 x\ln (x^2+9)\, \mathrm dx$ .

(Click for Solution)

Solution. Making the substitution $\displaystyle u = x^2 + 9 \Rightarrow \mathrm du = 2x\, \mathrm dx$ ,

$\displaystyle \begin{aligned} \int x\ln (x^2+9 )\, \mathrm dx &= \frac 12 \int \ln (x^2+9 ) \cdot 2x\, \mathrm dx \\ &= \frac 12 \int \ln u \, \mathrm du \\ &= \frac 12 (u \ln u - u) + C \\ &= \frac 12 \left((x^2 + 9) \ln (x^2 + 9) - (x^2 + 9) \right) + C. \end{aligned}$

Hence, substituting the limits,

$\displaystyle \begin{aligned} \int_1^2 x\ln (x^2+1)\, \mathrm dx &= \left[ \frac 12 \left( (x^2 + 9) \ln (x^2 + 9) - (x^2 + 9) \right) \right]_1^2 \\ &= \frac 12 \left( 13 \ln 13 - 13 \right) - \frac 12 \left( 10 \ln 10 - 10 \right) \\ &= \frac {13}2 \ln 13 - 5 \ln 10 - \frac 32.\end{aligned}$

Problem 13. Evaluate $\displaystyle \int \tan^3 x\, \mathrm dx$ .

(Click for Solution)

Solution. We solved this before, and solve it again for revision. Using the Pythagorean identity $\sec^2 x = \tan^2 x +1$ ,

$\displaystyle \begin{aligned} \int \tan^3 x\, \mathrm dx &= \int \tan x \tan^2 x\, \mathrm dx \\ &= \int \tan x (\sec^2 x - 1)\, \mathrm dx \\ &= \int \tan x \sec^2 x \, \mathrm dx - \int \tan x\, \mathrm dx \\ &= \int \tan x \sec^2 x \, \mathrm dx -(- \ln|{\cos x}|) + C.\end{aligned}$

For the first integral, make the substitution $\displaystyle u = \tan x \Rightarrow \mathrm dx = \frac{1}{\sec^2 x}\, \mathrm du$ . Then

$\displaystyle \begin{aligned}\int \tan x \sec^2 x \, \mathrm dx &= \int u \sec^2 x \cdot \frac{1}{\sec^2 x}\, \mathrm du \\ &= \int u\, \mathrm du \\ &= \frac 12 u^2 + C\\ &= \frac 12 \tan^2 x + C.\end{aligned}$

Therefore, $\displaystyle \int \tan^3 x\, \mathrm dx = \frac 12 \tan^2 x + \ln|{\cos x}| + C.$

Problem 14. Evaluate $\displaystyle \int \ln (x^2+9)\, \mathrm dx$ .

(Click for Solution)

Solution. Writing $\ln (x^2+9) = 1 \cdot \ln (x^2+9)$ and integrating by parts,

$\displaystyle \begin{aligned} \int \ln (x^2+9)\, \mathrm dx &= \underbrace{x}_{\mathrm I} \cdot \underbrace{\ln (x^2+9)}_{\mathrm S} - \int \underbrace{x}_{\mathrm I} \cdot \underbrace{\frac{2x}{x^2+9}}_{\mathrm D}\, \mathrm dx \\ &= x \ln(x^2 + 9) - \int \frac{2x^2}{x^2 + 9}\, \mathrm dx \\ &= x \ln(x^2 + 9) - 2\int \frac{(x^2 + 9) - 9}{x^2 + 1}\, \mathrm dx \\ &= x \ln(x^2 + 9) - 2\int \left( 1 - 9 \cdot \frac{1}{3^2 + x^2}\right) \, \mathrm dx \\ &= x \ln(x^2 + 9) - 2 \left( x - 9 \cdot \frac 1{3} \tan^{-1} \left( \frac x3 \right) \right) + C\\ &= x \ln(x^2 + 9) - 2x + 6\tan^{-1} \left( \frac x3 \right) + C.\end{aligned}$

Problem 15. Evaluate $\displaystyle \int \frac {2x+5}{x(x^2+9)}\, \mathrm dx$ .

(Click for Solution)

Solution. If we can decompose the integrand into

$\displaystyle \frac {2x+5}{x(x^2+9)} = \frac Ax + \frac{Bx+C}{x^2+9},$

Then a bit of basic integration yields

$\displaystyle \begin{aligned} \int \frac {2x+5}{x(x^2+9)}\, \mathrm dx &= \int \left(\frac Ax + \frac{Bx+C}{x^2+9}\right)\, \mathrm dx \\ &= \int \left(A \cdot \frac 1x + \frac B2 \cdot \frac{2x}{x^2+9} + C \cdot \frac 1{x^2 + 9} \right)\, \mathrm dx \\ &= A \int \frac 1x\, \mathrm dx + \frac B2 \int \frac {2x}{x^2 + 1}\, \mathrm dx + C \int \frac 1{3^2 + x^2}\, \mathrm dx \\ &= A \ln|x| + \frac B2 \ln (x^2 + 1) + C \cdot \frac 13 \tan^{-1} \left(\frac x3 \right) + D. \end{aligned}$

It remains to complete the decomposition. Clearing denominators,

$\begin{aligned}\frac {2x+5}{x(x^2+9)} &= \frac Ax + \frac{Bx+C}{x^2+9} \\ 2x+5 &= \frac {Ax(x^2+9)}x + \frac{(Bx+C)x(x^2+9)}{x^2+9} \\ 2x+5 &= A(x^2 + 9) + (Bx + C)x\\ 2x+5 &= (A+B)x^2 + Cx + 9A.\end{aligned}$

Setting $x = 0$ yields $A = 5/9$ .

Comparing coefficients yields $B = -5/9$ , and $C = 2$ .

Hence, by back-substituting,

$\displaystyle \begin{aligned} \int \frac 1{x(x^2+9)}\, \mathrm dx &= \frac 59 \ln|x| - \frac 5{18} \ln (x^2 + 9) + \frac 23 \tan^{-1} \left(\frac x3 \right) + C. \end{aligned}$

Problem 16. Evaluate $\displaystyle \int \frac {(2 \sin \theta +5)\cos \theta}{\sin \theta (\sin^2 \theta+9)}\, \mathrm d\theta$ .

(Click for Solution)

Solution. Making the substitution $\displaystyle u = \sin \theta \Rightarrow \mathrm d\theta = \frac{1}{\cos \theta}\, \mathrm du$ and applying the result from Problem 15,

$\displaystyle \begin{aligned} & \int \frac {(2 \sin \theta +5)\cos \theta}{\sin \theta (\sin^2 \theta+9)}\, \mathrm dx \\ &= \int \frac {(2u+5)\cos \theta}{u (u^2+9)} \cdot \frac{1}{\cos \theta}\, \mathrm du \\ &= \int \frac {2u+5}{u (u^2+9)} \, \mathrm du \\ &= \frac 59 \ln|u| - \frac 5{18} \ln (u^2 + 9) + \frac 23 \tan^{-1} \left(\frac u3 \right) + C \\ &= \frac 59 \ln| {\sin \theta} | - \frac 5{18} \ln ( \sin^2 \theta + 9 ) + \frac 23 \tan^{-1} \left(\frac {\sin \theta}3 \right) + C. \end{aligned}$

—Joel Kindiak, 13 Feb 25, 1757H

February 13, 2025

Calculus, math, mathematics, Physics, Probability
Several Differentiation Drills

For any integer $n$ , you may freely use the results

$\begin{aligned} (x^n)' &= nx^{n-1},\\ \sin'(x) &= \cos(x),\\ \cos'(x) &= -{\sin(x)},\\ \tan'(x) &= \sec^2(x), \\ (e^x)' &= e^x, \\ \ln'(x) &= \frac 1x, \end{aligned}$

from either previous posts or from real analysis. You may also apply the chain rule, product rule, and quotient rule however you wish.

Problem 1. For any $a > 0$ , prove that

$\displaystyle \begin{aligned} (a^x)' = a^x \ln(a), \quad \log_a'(x) = \frac{1}{x \ln(a)}. \end{aligned}$

(Click for Solution)

Solution. By exponential properties,

$a^x = (e^{\ln(a)})^x = e^{x \ln(a)}.$

Hence, by the chain rule,

$\displaystyle \begin{aligned} (a^x)' &= (e^{x \ln(a)})'\\ &= e^{x \ln(a)} \cdot \frac{\mathrm d}{\mathrm dx}(x \ln a) \\ &= e^{x \ln(a)} \cdot \ln(a) \\ &= a^x \ln(a). \end{aligned}$

By logarithm properties and linearity,

$\displaystyle \begin{aligned} \log_a'(x) &= \left( \frac{\ln(x)}{\ln(a)} \right)'\\ &= \frac{1}{\ln(a)} \ln'(x) \\ &= \frac{1}{\ln(a)} \cdot \frac 1x = \frac 1{x \ln (a)}. \end{aligned}$

Up till now, we have proven that $(x^n)' = nx^{n-1}$ only for integer values of $n$ . We extend this result to all real values of $n$ .

Problem 2. For any real $r$ , use Problem 1 to prove that $(x^r)' = rx^{r-1}$ .

(Click for Solution)

Solution. By exponential properties,

$x^r = (e^{\ln x})^r = e^{r \ln x}.$

Hence, by the chain rule,

$\displaystyle \begin{aligned} (x^r)' &= (e^{r \ln x})'\\ &= e^{r \ln x} \cdot \frac{\mathrm d}{\mathrm dx}(r \ln x) \\ &= x^r \cdot r \cdot \frac 1x = rx^{r-1}. \end{aligned}$

Example 1. By Problem 2,

$\displaystyle \frac{\mathrm d}{\mathrm dx}(x^{\sqrt 2}) = \sqrt 2 \cdot x^{\sqrt 2 - 1}, \quad \frac{\mathrm d}{\mathrm dx}(x^e) = e \cdot x^{e - 1}, \quad \frac{\mathrm d}{\mathrm dx}(x^\pi) = \pi \cdot x^{\pi - 1}.$

In particular,

$\displaystyle \frac{\mathrm d}{\mathrm dx}(x^e) = e \cdot x^{e - 1}\neq e^x = \frac{\mathrm d}{\mathrm dx}(e^x).$

Problem 3. Prove that

$\displaystyle \sec'(x) = \sec(x)\tan(x),\quad \csc'(x) = -{\csc(x)\cot(x)}.$

(Click for Solution)

Solution. Writing $\displaystyle \sec(x) = \frac 1{\cos(x)}$ , apply the quotient rule to obtain

$\displaystyle \begin{aligned} \sec'(x) &= \frac{ (1)' \cdot \cos(x) - 1 \cdot \cos'(x) }{ \cos^2(x) } \\ &= \frac{ 0 \cdot \cos(x) - 1 \cdot (-\sin(x)) }{ \cos^2(x) } \\ &= \frac{\sin(x)}{\cos^2(x)} = \frac{1}{\cos(x)} \cdot \frac{\sin(x)}{\cos(x)} = \sec(x) \tan(x). \end{aligned}$

We recall that $\csc(x) = \sec(\pi/2-x)$ so that

$\displaystyle \begin{aligned}\csc'(x) &= (\sec(\pi/2-x))' \\ &= \sec'(\pi/2-x)) \cdot (-1) \\ &= \sec(\pi/2-x) \tan(\pi/2-x) \cdot (-1) \\ &= \csc(x) \cot(x) \cdot (-1) \\ &= - \csc(x) \cot(x). \end{aligned}$

Problem 4. Given that $t > 0$ , $x \equiv x(t) > 0$ , and $y \equiv y(t) > 0$ , evaluate $\displaystyle \frac{\mathrm d}{\mathrm dt}(x^y)$ .

(Click for Solution)

Solution. Writing $x^y = e^{y \ln(x)}$ and using the chain rule,

$\begin{aligned} \frac{\mathrm d}{\mathrm dt}(x^y) &= \frac{\mathrm d}{\mathrm dt} (e^{y \ln(x)}) \\ &= e^{y \ln(x)} \left( \frac{\mathrm d}{\mathrm dt}(y \ln(x)) \right) \\ &= x^y \left( \frac{\mathrm dy}{\mathrm dt} \cdot \ln(x) + y \cdot \frac{\mathrm d}{\mathrm dt} ( \ln(x)) \right) \\ &= x^y \left( \frac{\mathrm dy}{\mathrm dt} \cdot \ln(x) + y \cdot \frac 1x \cdot \frac{\mathrm dx}{\mathrm dt} \right) \\ &= x^y \ln(x) \cdot \frac{\mathrm dy}{\mathrm dt} + yx^{y-1} \cdot \frac{\mathrm dx}{\mathrm dt}. \end{aligned}$

Remark 1. If $x = t$ and $y$ is constant, we recover the power rule. If $x$ is constant and $y = t$ , we recover Problem 1. Furthermore, if $y = x = t$ , we obtain

$\displaystyle \frac{\mathrm d}{\mathrm dx}(x^x) = x^x (\ln(x) + 1).$

In the language of partial derivatives, denoting $z = x^y \equiv { x(t) }^{ y(t) }$ ,

$\displaystyle \frac{\mathrm d z}{\mathrm dt} = \frac{\partial z}{\partial y} \cdot \frac{\mathrm dy}{\mathrm dt} + \frac{\partial z}{\partial x} \cdot \frac{\mathrm dx}{\mathrm dt}.$

Problem 5. Suppose $f, f^{-1}$ are differentiable on $\mathbb R$ . Evaluate $(f^{-1})'(x)$ .

(Click for Solution)

Solution. By definition of the inverse function,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx} (x) &= \frac{\mathrm d}{\mathrm dx} (f(f^{-1}(x))) \\ 1 &= f'(f^{-1}(x)) \cdot \frac{\mathrm d}{\mathrm dx}(f^{-1}(x)) \\ &= f'(f^{-1}(x)) \cdot f'(f^{-1}(x)) \\ (f^{-1})'(x) &= \frac 1{f'(f^{-1}(x))}. \end{aligned}$

—Joel Kindiak, 9 Feb 25, 2126H

February 9, 2025

ai, artificial-intelligence, math, mathematics, Physics
Several Limit Drills

Problem 1. For any real number $a$ , evaluate the limit

$\displaystyle \lim_{h \to 0} \frac{(a+h)^2 - a^2}{h}.$

(Click for Solution)

Solution. Expanding the numerator yields

$\displaystyle \begin{aligned} \frac{(a+h)^2 - a^2}{h} &= \frac{a^2 + 2ah + h^2 - a^2}{h} \\ &= \frac{2ah + h^2}{h}\\ &= \frac{(2a+h)h}{h}\\ &= 2a + h. \end{aligned}$

Taking $h \to 0$ yields

$\displaystyle \begin{aligned} \lim_{h \to 0} \frac{(a+h)^2 - a^2}{h} &=\lim_{h \to 0} (2a+h) \\ &= 2a+0 \\ &= 2a. \end{aligned}$

Problem 2. For any real number $a > 0$ , evaluate the limit

$\displaystyle \lim_{h \to 0} \frac{\sqrt{a+h} - \sqrt{a}}{h}.$

(Click for Solution)

Solution. Rationalising the numerator,

$\displaystyle \begin{aligned} \frac{\sqrt{a+h} - \sqrt{a}}{h} &= \frac{\sqrt{a+h} - \sqrt{a}}{h} \cdot \frac{\sqrt{a+h} + \sqrt{a}}{\sqrt{a+h} + \sqrt{a}} \\ &= \frac{(\sqrt{a+h})^2 - (\sqrt{a})^2}{h(\sqrt{a+h} + \sqrt{a})} \\ &= \frac{(a+h) - a}{h(\sqrt{a+h} + \sqrt{a})} \\ &= \frac{h}{h(\sqrt{a+h} + \sqrt{a})} \\ &= \frac{1}{\sqrt{a+h} + \sqrt{a}}.\end{aligned}$

Taking $h \to 0$ yields

$\displaystyle \begin{aligned} \lim_{h \to 0} \frac{(a+h)^2 - a^2}{h} &=\lim_{h \to 0} \frac{1}{\sqrt{a+h} + \sqrt{a}} \\ &= \frac{1}{\sqrt{a + 0} + \sqrt a} \\ &= \frac{1}{2\sqrt a}. \end{aligned}$

Problem 3. For any real number $a \neq 0$ , evaluate the limit

$\displaystyle \lim_{h \to 0} \frac{\frac 1{a+h} - \frac 1a}{h}.$

(Click for Solution)

Solution. Combining denominators,

$\displaystyle \begin{aligned} \frac{\frac 1{a+h} - \frac 1a}{h} &= \frac{1}{h} \left(\frac 1{a+h} - \frac 1a\right) \\ &= \frac{1}{h} \cdot \frac{a - (a+h)}{a(a+h)} \\ &= \frac{1}{h} \cdot \frac{-h}{a(a+h)} \\ &= - \frac{1}{a(a+h)}.\end{aligned}$

Taking $h \to 0$ yields

$\displaystyle \begin{aligned} \lim_{h \to 0} \frac{\frac 1{a+h} - \frac 1a}{h} &=\lim_{h \to 0} - \frac{1}{a(a+h)} \\ &= - \frac{1}{a(a+0)} \\ &= -\frac{1}{a^2}. \end{aligned}$

Remark 1. The derivative $f'$ of a function $f$ is defined by

$f'(a) := \displaystyle \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}$

whenever the right-hand side exists. In this exercise, we applied this definition to the functions $f(x) = x^2$ , $f(x) = \sqrt x$ , $f(x) = 1/x$ , and evaluated their derivatives respectively:

$\displaystyle (x^2)' = 2x,\quad (\sqrt x)' = \frac{1}{2\sqrt x},\quad \left(\frac 1x\right)' = -\frac 1{x^2}.$

—Joel Kindiak, 9 Feb 25, 1116H

February 9, 2025

Calculus, machine-learning, math, mathematics, Physics
Some Integration Drills

Problem 1. Evaluate the integral $\displaystyle \int(x^2 + \tan x - 3)\, \mathrm dx$ .

(Click for Solution)

Solution. By the linearity of integration,

$\displaystyle \begin{aligned}\int(x^2 + \tan x - 3)\, \mathrm dx &= \int x^2\, \mathrm dx + \int \tan x \, \mathrm dx - \int 3\, \mathrm dx \\ &= \frac 13 x^3 - \ln|{\cos x}| - 3x + C.\end{aligned}$

Problem 2. Evaluate the integral $\displaystyle \int \left(\frac{1}{9 + x^2} + \frac{1}{\sqrt{4-x^2}} \right)\, \mathrm dx$ .

(Click for Solution)

Solution. By the linearity of integration,

$\displaystyle \begin{aligned} \int \left(\frac{1}{9 + x^2} + \frac{1}{\sqrt{4-x^2}} \right)\, \mathrm dx &= \int \frac{1}{9 + x^2} \, \mathrm dx + \int \frac{1}{ \sqrt{4-x^2} } \, \mathrm dx \\ &= \int \frac{1}{3^2 + x^2} \, \mathrm dx + \int \frac{1}{ \sqrt{2^2-x^2} } \, \mathrm dx \\ &= \frac 13 \tan^{-1} \left( \frac x3 \right) + \sin^{-1} \left( \frac x2 \right) + C. \end{aligned}$

Problem 3. Evaluate the integral $\displaystyle \int_0^1 x \sec(x^2) \tan(x^2)\, \mathrm dx$ .

(Click for Solution)

Solution. By the substitution $u = x^2 \Rightarrow \frac 12 \mathrm du = x\, \mathrm dx$ ,

$\displaystyle \begin{aligned} \int x \sec(x^2) \tan(x^2)\, \mathrm dx &= \int \sec(x^2) \tan(x^2) \cdot x\, \mathrm dx \\ &= \int \sec(u) \tan(u) \cdot \frac{1}{2}\, \mathrm du \\ &= \frac 12 \int \sec(u) \tan(u)\, \mathrm du \\ &= \frac 12 \sec(u) + C \\ &= \frac 12 \sec(x^2) + C. \end{aligned}$

Hence, substituting the limits,

$\displaystyle \begin{aligned} \int_0^1 x \sec(x^2) \tan(x^2)\, \mathrm dx &= \frac 12 [\sec(x^2)]_0^1\\ &= \frac 12 (\sec(1) - \sec(0)) \\ &= \frac 1{2 \cos(1)} - \frac 12 \\ &\approx 0.425, \end{aligned}$

when approximated to 3 decimal places.

Problem 4. Evaluate the integral $\displaystyle \int \tan^3 x\, \mathrm dx$ .

(Click for Solution)

Solution. Using the Pythagorean identity $\sec^2 x = \tan^2 x +1$ ,

$\displaystyle \begin{aligned} \int \tan^3 x\, \mathrm dx &= \int \tan x \tan^2 x\, \mathrm dx \\ &= \int \tan x (\sec^2 x - 1)\, \mathrm dx \\ &= \int \tan x \sec^2 x \, \mathrm dx - \int \tan x\, \mathrm dx \\ &= \int \tan x \sec^2 x \, \mathrm dx -(- \ln|{\cos x}|) + C.\end{aligned}$

For the first integral, make the substitution

$\displaystyle u = \tan x \quad \Rightarrow \quad \mathrm du = \sec^2 x\, \mathrm dx.$

Then

$\displaystyle \begin{aligned}\int \tan x \sec^2 x \, \mathrm dx &= \int u \, \mathrm du \\ &= \frac 12 u^2 + C\\ &= \frac 12 \tan^2 x + C.\end{aligned}$

Therefore, $\displaystyle \int \tan^3 x\, \mathrm dx = \frac 12 \tan^2 x + \ln|{\cos x}| + C.$

Problem 5. Evaluate the integral $\displaystyle \int \sin(\ln \theta)\, \mathrm d\theta$ .

(Click for Solution)

Solution. Make the substitution $u = \ln \theta \Rightarrow \mathrm d\theta = e^u\, \mathrm du$ . Then

$\displaystyle \int \sin(\ln \theta)\, \mathrm d\theta = \int e^u \sin(u)\, \mathrm du.$

Integrating by parts twice,

$\displaystyle \begin{aligned}\int e^u \sin(u)\, \mathrm du &= \underbrace{e^u}_{\mathrm I}\, \underbrace{\sin(u)}_{\mathrm S} - \int \underbrace{e^u}_{\mathrm I}\, \underbrace{\cos(u)}_{\mathrm D}\, \mathrm du \\ &= e^u \sin(u) - \int e^u \cos(u)\, \mathrm du\\ &= e^u \sin(u) - \left(\underbrace{e^u}_{\mathrm I}\, \underbrace{\cos(u)}_{\mathrm S} - \int \underbrace{e^u}_{\mathrm I}\, \underbrace{-\sin(u)}_{\mathrm D}\, \mathrm du\right) \\ &= e^u \sin(u) - \left(e^u \cos(u) + \int e^u \sin(u) \, \mathrm du\right) \\ &= e^u \sin(u) - e^u \cos(u) - \int e^u \sin(u)\, \mathrm du \\&= e^u (\sin(u) - \cos(u)) - \int e^u \sin(u)\, \mathrm du.\end{aligned}$

By algebruh,

$\displaystyle \begin{aligned} \int e^u \sin(u)\, \mathrm du + \int e^u \sin(u)\, \mathrm du &= e^u (\sin(u) - \cos(u)) \\ 2 \int e^u \sin(u)\, \mathrm du &= e^u (\sin(u) - \cos(u)) \\ \int e^u \sin(u)\, \mathrm du &= \frac 12 e^u (\sin(u) - \cos(u)) + C\\ \int \sin(\ln \theta)\, \mathrm d\theta &= \frac 12 \theta (\sin(\ln \theta) - \cos(\ln \theta)) + C.\end{aligned}$

Problem 6. Evaluate the integral $\displaystyle \int \left(\frac{1}{36 - x^2} + \frac{1}{\sqrt{36-x^2}} \right)\, \mathrm dx$ .

(Click for Solution)

Solution. By the linearity of integration,

$\displaystyle \begin{aligned} \int \left(\frac{1}{36 - x^2} + \frac{1}{\sqrt{36 -x^2}} \right)\, \mathrm dx &= \int \frac{1}{36 - x^2} \, \mathrm dx + \int \frac{1}{ \sqrt{36-x^2} } \, \mathrm dx \\ &= \int \frac{1}{6^2 - x^2} \, \mathrm dx + \int \frac{1}{ \sqrt{6^2-x^2} } \, \mathrm dx \\ &= \frac 1{2 \cdot 6} \ln \left| \frac{6+x}{6-x} \right| + \sin^{-1} \left( \frac x6 \right) + C\\ &= \frac 1{12} \ln \left| \frac{6+x}{6-x} \right| + \sin^{-1} \left( \frac x6 \right) + C. \end{aligned}$

Problem 7. Evaluate the integral $\displaystyle \int_0^2 x \sin(x^2 + 5)\, \mathrm dx$ .

(Click for Solution)

Solution. Making the substitution $u = x^2 + 5$ yields

$\frac 12 \mathrm du = x\, \mathrm dx,\quad u(0) = 5, \quad u(2) = 9.$

Therefore,

$\displaystyle \begin{aligned} \int_0^2 x \sin(x^2 + 5)\, \mathrm dx &= \int_0^2 \sin(x^2 + 5) \cdot x\, \mathrm dx \\ &= \int_5^9 \sin(u) \cdot \frac 12\, \mathrm du \\ &= \frac 1{2}\int_5^9 \sin(u)\, \mathrm du \\ &= \frac 12 [-{\cos(u)}]_5^9 \\ &= \frac 12 (-{\cos(9)} - (-{\cos(5)})) \\ &= \frac 12 (\cos(5) - \cos(9)) \approx 0.597, \end{aligned}$

when approximated to 3 decimal places.

Problem 8. Evaluate the integral $\displaystyle \int \cot^3 x\, \mathrm dx$ .

(Click for Solution)

Solution. Recall that $\cot(x) = \tan(\pi/2 - x)$ . Making the substitution $u = \pi/2-x \Rightarrow \mathrm du = -\mathrm dx$ then using the result in Problem 4,

$\displaystyle \begin{aligned} \int \cot^3 x\, \mathrm dx &= \int \tan^3 (\pi/2-x)\, \mathrm dx \\ &= \int \tan^3(u) \cdot -1\, \mathrm du \\ &= -\int \tan^3(u)\, \mathrm du \\ &= -\left( \frac 12 \tan^2 u + \ln|{\cos u}| \right) + C \\ &= -\left( \frac 12 \tan^2 (\pi/2-x) + \ln|{\cos (\pi/2-x)}| \right) + C \\ &= -\left( \frac 12 \cot^2 x + \ln|{\sin x}| \right) + C \\ &= - \frac 12 \cot^2 x - \ln|{\sin x}| + C \end{aligned}$

Problem 9. Evaluate the integral $\displaystyle \int e^{2\theta} \sin (e^\theta) \, \mathrm d\theta$ .

(Click for Solution)

Solution. Making the substitution $\displaystyle u = e^{\theta} \Rightarrow \mathrm d\theta = \frac 1{u}\, \mathrm du$ then integrating by parts,

$\displaystyle \begin{aligned} \int e^{2\theta} \sin(e^\theta) \, \mathrm d\theta &= \int u^2 \sin(u)\cdot \frac 1{u}\, \mathrm du \\ &= \int u \sin(u)\, \mathrm du\\ &= \underbrace{-\cos(u)}_{\mathrm I} \underbrace{u}_{\mathrm S} - \int \underbrace{-\cos(u)}_{\mathrm I}\underbrace{1}_{\mathrm D}\, \mathrm du \\ &= -u \cos u + \int \cos u\, \mathrm du\\ &= -u \cos u + \sin u + C\\ &= -e^\theta \cos(e^\theta) + \sin(e^\theta) + C. \end{aligned}$

Problem 10. Given that

$f(x) = \begin{cases} 1 - |x|, & |x| \leq 1, \\ 0, &\text{otherwise},\end{cases}$

evaluate $\displaystyle \int_0^{2} f(x^2-1)\, \mathrm dx$ .

(Click for Solution)

Solution. By the definition of $f$ ,

$f(x^2 - 1) = \begin{cases} 1 - |x^2 - 1|, & |x^2 - 1| \leq 1, \\ 0, &\text{otherwise}.\end{cases}$

Parsing the inequality $|x^2 - 1| \leq 1 \iff |x| \leq \sqrt 2$ ,

$f(x^2 - 1) = \begin{cases} 1 - |x^2 - 1|, & |x| \leq \sqrt 2, \\ 0, &\text{otherwise}.\end{cases}$

By definition of the absolute value,

$|x^2 - 1| = \begin{cases} x^2 - 1, & |x| \geq 1, \\ 1 - x^2, &|x| \leq 1.\end{cases}$

Parsing the inequalities again,

$f(x^2 - 1) = \begin{cases} x^2, & 0 \leq |x| \leq 1, \\ 2-x^2 , & 1 \leq |x| \leq \sqrt 2, \\ 0, &\text{otherwise}.\end{cases}$

Therefore, we can evaluate the integral as

$\begin{aligned} \displaystyle \int_0^{2} f(x^2-1)\, \mathrm dx &= \int_0^1 f(x^2 - 1)\, \mathrm dx + \int_1^{\sqrt 2} f(x^2 - 1)\, \mathrm dx + \int_{\sqrt 2}^{2} 0\, \mathrm dx \\ &= \int_0^1 x^2\, \mathrm dx + \int_1^{\sqrt 2} (2 - x^2)\, \mathrm dx+ 0 \\ &= \left[ \frac{x^3}{3}\right]_0^1 + \left[ 2x - \frac{x^3}{3} \right]_1^{\sqrt 2} \\ &= \left( \frac 13 - 0 \right) + \left( 2\sqrt 2 - \frac{2\sqrt 2}{3} \right) \\ &= \frac 13 + \frac{4}{3}\sqrt 2. \end{aligned}$

—Joel Kindiak, 20 Jan 25, 0925H

January 20, 2025

ai, Calculus, math, mathematics, Physics
When Integrals Approximate One

In the following problems, let $f$ be a function that is continuous on $\mathbb R$ .

Problem 1. Given that $\displaystyle \lim_{x \to \infty} \frac{1}{x} \int_0^x f(t)\, \mathrm dt = 1$ , prove that $\displaystyle \lim_{x \to \infty} f(x) = 1$ if the limit exists.

(Click for Solution)

Solution. We first claim that $\displaystyle \int_0^x f(t)\, \mathrm dt \to \infty$ . Fix $M > 0$ . By hypothesis, for any $\epsilon > 0$ , there exists $K > 0$ such that $x > K$ implies

$\displaystyle \left|\frac{1}{x} \int_0^x f(t)\, \mathrm dt - 1\right| < \epsilon \quad \Rightarrow \quad \int_0^x f(t)\, \mathrm dt > (1 - \epsilon)x.$

Setting $\epsilon = 1/2$ , choose $N := \max\{K, 2M\}$ so that $x > N$ implies

$\displaystyle \int_0^x f(t)\, \mathrm dt > (1-1/2) \cdot 2M = M.$

Therefore $\displaystyle \int_0^x f(t)\, \mathrm dt \to \infty$ . By L’Hôpital’s rule and the fundamental theorem of calculus,

$\displaystyle \lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{\frac{\mathrm d}{\mathrm dx} \int_0^x f(t)\, \mathrm dt}{\frac{\mathrm d}{\mathrm dx} (x)} = \lim_{x \to \infty} \frac{\int_0^x f(t)\, \mathrm dt}{x} = 1.$

But what if we don’t know that the limit exists? The added feature of $f$ being increasing helps.

Problem 2. Given that $\displaystyle \lim_{x \to \infty} \frac{1}{x} \int_0^x f(t)\, \mathrm dt = 1$ , prove that $\displaystyle \lim_{x \to \infty} f(x) = 1$ if $f$ is increasing.

(Click for Solution)

Solution. Fix $\epsilon > 0$ . By the given condition, for any $k > 0$ to be chosen, there exists $K > 0$ such that for $x > K$ ,

$\displaystyle -k \epsilon < \frac{1}{x} \int_0^x f(t)\, \mathrm dt - 1 < k \epsilon.$

We claim that $f(x) \leq 1$ for any $x$ . Suppose $f(N) > 1$ for some $N$ . Define $\epsilon_0 := (f(N)-1)/2$ so that for $x > N$ , $f(x) \geq f(N) > 1 + \epsilon_0$ . Then

$\displaystyle \begin{aligned} \frac{1}{x} \int_0^x f(t)\, \mathrm dt &= \frac{1}{x} \int_0^N f(t)\, \mathrm dt + \frac{1}{x} \int_N^x f(t)\, \mathrm dt \\ &\geq \frac{1}{x} \int_0^N f(t)\, \mathrm dt + (1+\epsilon_0) (1-N/x) \\ &\geq \frac{N}{x} \cdot (f(0) - (1+\epsilon_0)) + (1+\epsilon_0) . \end{aligned}$

Taking $x \to \infty$ yields

$\displaystyle 1 = \lim_{x \to \infty} \frac{1}{x} \int_0^x f(t)\, \mathrm dt \geq 0 \cdot (f(0) - (1+\epsilon_0)) + (1+\epsilon_0) > 1,$

a contradiction. Therefore, $f(x) \leq 1$ for any $x$ . On the other hand, apply the mean value theorem and the fundamental theorem of calculus to find $c \in (0, x)$ such that

$\displaystyle f(c) = \frac{1}{x} \int_0^x f(t)\, \mathrm dt > 1 - k\epsilon.$

Setting $k = 1$ and using the fact that $f$ is increasing,

$1 - \epsilon < f(c) \leq f(x) \leq 1 < 1 +\epsilon\quad \Rightarrow \quad |f(x) - 1| < \epsilon.$

It turns out that we can we relax our hypothesis. However, this will require more rudimentary ideas.

Problem 3. Given that $\displaystyle \lim_{x \to \infty} \frac{1}{x} \int_0^x f(t)\, \mathrm dt = 1$ , prove that $\displaystyle \lim_{x \to \infty} f(x) = 1$ .

(Click for Solution)

Solution. Fix $\epsilon > 0$ . By the given condition, for any $k > 0$ to be chosen, there exists $K > 0$ such that for $x > K$ ,

$\displaystyle -k \epsilon < \frac{1}{x} \int_0^x f(t)\, \mathrm dt - 1 < k \epsilon.$

By algebra, since $x > K > 0$ ,

$\displaystyle -k \epsilon x < \int_0^x f(t)\, \mathrm dt - x < k \epsilon x.$

In particular, the inequality holds replacing $x$ with $x+h$ for $|h| < (x-K)/2$ so that

$\displaystyle -k \epsilon (x+h) < \int_0^{x+h} f(t)\, \mathrm dt - (x+h) < k \epsilon (x+h).$

Subtracting the inequalities then dividing by $h$ ,

$\displaystyle -k \epsilon < \frac 1h \int_x^{x+h} f(t)\, \mathrm dt - 1 < k \epsilon.$

Taking $h \to 0$ and applying the fundamental theorem of calculus,

$\displaystyle -k \epsilon \leq f(x)-1 \leq k \epsilon.$

Setting $k = 1/2$ ,

$\displaystyle -\epsilon < -k \epsilon \leq f(x)-1 \leq k \epsilon < \epsilon\quad \Rightarrow \quad |f(x) - 1| < \epsilon.$

By the $\epsilon$ – $K$ definition for limits, $\displaystyle \lim_{x \to \infty} f(x) = 1$ .

—Joel Kindiak, 26 Nov 24, 0110H

November 26, 2024

Calculus, machine-learning, math, mathematics, Physics

Category: Calculus

Limits

Differentiation

Integration