University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 2 - Section 2.2 - Limit of a Function and Limit Laws - Exercises - Page 67: 57

Answer

$$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=2$$

Work Step by Step

$$f(x)=x^2\hspace{1cm} x=1$$ So $f(x+h)=(x+h)^2=x^2+2xh+h^2$ Therefore, $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{(x^2+2xh+h^2)-(x^2)}{h}$$ $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{2xh+h^2}{h}$$ $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}(2x+h)$$ Substitute $x=1$ here: $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}(2+h)$$ $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=2+0=2$$
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