## University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson

# Chapter 2 - Section 2.2 - Limit of a Function and Limit Laws - Exercises - Page 67: 61

#### Answer

$$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\frac{\sqrt7}{14}$$

#### Work Step by Step

$$f(x)=\sqrt x\hspace{1cm} x=7$$ So $f(x+h)=\sqrt{x+h}$ Therefore, $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt x}{h}$$ -Multiply both numerator and denominator with $\sqrt{x+h}+\sqrt x$, so that - Numerator: $(\sqrt{x+h}-\sqrt x)(\sqrt{x+h}+\sqrt x)=(x+h)-x=h$ - Denominator: $h(\sqrt{x+h}+\sqrt x)$ That means $$\lim_{h\to0}\frac{f(x+h)f(x)}{h}=\lim_{h\to0}\frac{h}{h(\sqrt{x+h}+\sqrt x)}$$ $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{1}{(\sqrt{x+h}+\sqrt x)}$$ - Substitute $x=7$ here: $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{1}{\sqrt{7+h}+\sqrt7}$$ $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\frac{1}{\sqrt{7+0}+\sqrt7}=\frac{1}{\sqrt7+\sqrt7}$$ $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\frac{1}{2\sqrt7}$$ Multiply both numerator and denominatory with $\sqrt7$: $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\frac{\sqrt7}{2\times7}=\frac{\sqrt7}{14}$$

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