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: 35


The graph of $f$ has a tangent at the origin.

Work Step by Step

$f(x)=x^2\sin(1/x)$ for $x\ne0$ and $f(x)=0$ for $x=0$ For the graph of $f$ to have a tangent at the origin, the value of its slope there, the derivative $f'(0)$, must exist. In other words, we will see if the formula for $f'(0)$ exists or not. In detail, $$f'(0)=\lim_{h\to0}\frac{f(h+0)-f(0)}{h}=\lim_{h\to0}\frac{f(h)-f(0)}{h}$$ Remember here that - For $h\to0$, $h$ does not equal $0$, so $f(h)=h^2\sin(1/h)$ - For $f(0)$, having $x=0$ means $f(0)=0$ $$f'(0)=\lim_{h\to0}\frac{h^2\sin(1/h)-0}{h}=\lim_{h\to0}\frac{h^2\sin(1/h)}{h}$$ $$f'(0)=\lim_{h\to0}\Big(h\sin(1/h)\Big)$$ It is known that $-1\le\sin(1/h)\le1$ So, $-h\le h\sin(1/h)\le h$ Because $\lim_{h\to0}h=\lim_{h\to0}-h=0$, according to the Sandwich Theorem, $$f'(0)=\lim_{h\to0}\Big(h\sin(1/h)\Big)=0$$ In other words, $f'(0)$ does exist, and so does the slope of the tangent to $f$ at the origin. Therefore, the graph of $f$ has a tangent at the origin.
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