University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.1 - Tangents and the Derivative at a Point - Exercises - Page 118: 22


The slope of the curve at $x=0$ is $2$.

Work Step by Step

The slope $m$ of the tangent line of the curve $f(x)$ at $A(a,b)$ is also the slope of the curve at that point, which is calculated by $$m=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$$ $$f(x)=\frac{x-1}{x+1}\hspace{1cm}x=0$$ To calculate the slope of the curve at $x=0$, we apply the above formula: $$m=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$$ Here $a=x=0$ $$m=\lim_{h\to0}\frac{f(h+0)-f(0)}{h}=\lim_{h\to0}\frac{f(h)-f(0)}{h}$$ $$m=\lim_{h\to0}\frac{\frac{h-1}{h+1}-\frac{0-1}{0+1}}{h}=\lim_{h\to0}\frac{\frac{h-1}{h+1}+1}{h}=\lim_{h\to0}\frac{\frac{h-1+h+1}{h+1}}{h}$$ $$m=\lim_{h\to0}\frac{2h}{h(h+1)}=\lim_{h\to0}\frac{2}{h+1}$$ $$m=\frac{2}{0+1}=2$$
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