KindiakMath

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 25H, 1116H

,
, , , ,

Published by

joelkindiak

Leave a comment