KindiakMath

Tricky Limits

Big Idea

We write \displaystyle \lim_{x \to c} f(x) = L to mean that f(x) \to L as x \to c. In particular, if g is continuous at c and f(x) = g(x) for x \neq c, then

\displaystyle \lim_{x \to c} f(x) = \lim_{x \to c} g(x) = g(c).

Furthermore, the usual limit laws hold.

Questions

Question 1. Evaluate the limit \displaystyle \lim_{x \to 2} \frac{x^2 - 4}{x^3 - 8}.

(Click for Solution)

Solution. Since 2 is a root of x^3 - 8, x-2 is a factor of x^3 - 8:

\begin{aligned} x^3 - 8 &= (x-2)(x^2 + kx + 4) \\ &= x^3 + (k-2)x^2 +(4-2k)x - 8. \end{aligned}

Comparing coefficients, k = 2. Hence,

\begin{aligned} \lim_{x \to 2} \frac{x^2 - 4}{x^3 - 8} &= \lim_{x \to 2} \frac{(x-2)(x+2)}{(x-2)(x^2 + 2x + 4)} \\ &= \lim_{x \to 2} \frac{x+2}{x^2 + 2x + 4} \\ &= \frac{2+2}{2^2 + 2\cdot 2 + 4} = \frac 13.\end{aligned}

Alternate Solution. Make the substitution x = 2 +t, so that x \to 2 leads to t \to 2. Then

\begin{aligned} \lim_{x \to 2} \frac{x^2 - 4}{x^3 - 8} &= \lim_{t \to 0} \frac{(2+t)^2 - 4}{(2+t)^3 - 8} \\ &= \lim_{t \to 0} \frac{ (4 + 4t + t^2) - 4}{ (8 + 12t + 6t + t^3) - 8} \\ &= \lim_{t \to 0} \frac{ 4t + t^2 }{ 12t + 6t^2 + t^3 } \\ &= \lim_{t \to 0} \frac{ 4 + t }{ 12 + 6t + t^2 } \\ &= \frac{ 4 + 0 }{ 12 + 6 \cdot 0 + 0^2 } = \frac 13.\end{aligned}

Question 2. Evaluate the limit \displaystyle \lim_{x \to 4} \frac{\sqrt{1+2x} - 3}{\sqrt x-2}.

(Click for Solution)

Solution. Rationalising both numerator and denominator,

\begin{aligned} \lim_{x \to 4} \frac{\sqrt{1+2x} - 3}{\sqrt x-2} &= \lim_{x \to 4} \left( (\sqrt{1+2x} - 3) \cdot \frac{1}{\sqrt x-2} \right) \\ &= \lim_{x \to 4} \left( (\sqrt{1+2x} - 3) \cdot \frac{\sqrt{1+2x} + 3}{\sqrt{1+2x} + 3} \cdot \frac{1}{\sqrt x-2} \cdot \frac{\sqrt x + 2}{\sqrt x + 2} \right) \\ &= \lim_{x \to 4} \left( \frac{(1+2x) - 9}{\sqrt{1+2x} + 3} \cdot \frac{\sqrt x + 2}{x-4} \right) \\ &= \lim_{x \to 4} \left( \frac{ 2(x-4) }{\sqrt{1+2x} + 3} \cdot \frac{\sqrt x + 2}{x-4} \right) \\ &= 2\cdot \lim_{x \to 4} \frac{ \sqrt x + 2 }{\sqrt{1+2x} + 3} = 2 \cdot \frac{\sqrt 4+2}{\sqrt{1 + 2 \cdot 4} + 3} = \frac 43. \end{aligned}

Question 3. Evaluate the limit \displaystyle \lim_{x \to \infty} \frac{\sqrt{4x^2 + 5}}{\sqrt[3]{x^3 - 1}}.

(Click for Solution)

Solution. By factoring x = \sqrt{x^2} = \sqrt[3]{x^3} on both the numerator and the denominator,

\begin{aligned} \lim_{x \to \infty} \frac{\sqrt{4x^2 + 5}}{\sqrt[3]{x^3 - 1}} &= \lim_{x \to \infty} \frac{\sqrt{x^2 \cdot (4 + 5 / x^2)}}{\sqrt[3]{x^3 \cdot (1 - 1/ x^3)}} \\ &= \lim_{x \to \infty} \frac{\sqrt{x^2} \cdot \sqrt{ 4 + 5 / x^2 }}{\sqrt[3]{x^3} \cdot \sqrt[3]{ 1 - 1/x^3 } } \\ &= \lim_{x \to \infty} \frac{x \cdot \sqrt{ 4 + 5/ x^{2} }}{x \cdot \sqrt[3]{ 1 - 1/x^{3} } } \\&= \lim_{x \to \infty} \frac{\sqrt{ 4 + 5 /x^{2} }}{\sqrt[3]{ 1 - 1/x^{3} } } = \frac{\sqrt{ 4 + 0 }}{\sqrt[3]{ 1 - 0 } } = 2. \end{aligned}

—Joel Kindiak, 3 Sept 25, 1804H

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