Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 2: Limits and Continuity - Section 2.2 - Limit of a Function and Limit Laws - Exercises 2.2 - Page 56: 22



Work Step by Step

We want to find $\lim\limits_{h \to 0}\dfrac{\sqrt{5h+4}-2}{h}$, but if we try to substitute $h=0$ directly into our function, we get zero in both the numerator and the denominator, and we can't use the quotient rule for limits because the limit of the denominator as $h$ approaches zero is zero. But, we can obtain a factor of $h$ in the numerator by multiplying the numerator and denominator by $\sqrt{5h+4}+2$: $\dfrac{\sqrt{5h+4}-2}{h}=\dfrac{\sqrt{5h+4}-2}{h}\cdot \dfrac{\sqrt{5h+4}+2}{\sqrt{5h+4}+2}=\dfrac{5h+4-4}{h(\sqrt{5h+4}+2)}=\dfrac{5h}{h(\sqrt{5h+4}+2)}.$ Note that $\dfrac{5h}{h(\sqrt{5h+4}+2)}=\dfrac{5}{\sqrt{5h+4}+2}$ for all $h \neq 0.$ Thus $\dfrac{\sqrt{5h+4}-2}{h}=\dfrac{5}{\sqrt{5h+4}+2}$ for all $h \neq 0.$ This means $\lim\limits_{h \to 0}\dfrac{\sqrt{5h+4}-2}{h}=\lim\limits_{h \to 0}\dfrac{5}{(\sqrt{5h+4}+2)}=\dfrac{5}{\sqrt{5(0)+4}+2}=\dfrac{5}{\sqrt{4}+2}=\dfrac{5}{2+2}=\dfrac{5}{4}.$
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