## Calculus: Early Transcendentals 8th Edition

a) let $x=\frac{1}{k\pi}$ where k is an integer $f(\frac{1}{k\pi})=\tan\frac{1}{\frac{1}{k\pi}}=\tan(k\pi)=0$ b) let $x=\frac{4}{(4k+1)\pi}$ where k is an integer $f(\frac{4}{(4k+1)\pi})=\tan\frac{1}{\frac{4}{(4k+1)\pi}}=\tan\frac{(4k+1)\pi}{4}=\tan(\frac{\pi}{4}+k\pi)=1$ c) From previous parts, it is evident that $\tan\frac{1}{x}$ oscillates between 0 and 1 infinitely as $x$ approaches 0 from the right. Since $f(x)$ does not approach a fixed value as x approaches 0 from the right, $\lim\limits_{x \to 0^+}\tan\frac{1}{x}$ does not exist.