Answer
(a)
${tan(nx)}$=$0$. (Remember that the tangent function has
period $\pi$.)
(b)
${( tan \frac{\pi}{4})}$=$1$
(c)
Thus,$ {\lim\limits_{x \to 0} (tan\frac{1}{x})}$ does not exist since $f(x)$ does not get close to a fixed number as $x$ →$0$
Work Step by Step
(a)
For any positive integer $n$, if $x$ =${ \frac{1}{n\pi}}$
then$ f(x)$ = ${tan\frac{1}{x}}$=${tan(nx)}$=$0$. (Remember that the tangent function has
period $\pi$.)
(b)
For any nonnegative number $n$, if $x$ = ${\frac{4}{(4n+1)\pi}}$ ,
then
$f(x)$ =$( tan\frac{1}{x})$=$( tan\frac{(4n+1)\pi}{4})$ = ${tan(\frac{4n\pi}{4}+\frac{\pi}{4})}$=${ tan(n\pi+\frac{\pi}{4})}$ =${( tan \frac{\pi}{4})}$=$1$
(c)
From part (a),
$f(x)$=$0$ infinitely often as $x$ → $0$.
From part (b)
,$f(x)$=$1$ infinitely often as $x$ →$0$ Thus,$ {\lim\limits_{x \to 0} (tan\frac{1}{x})}$ does not exist since $f(x)$ does not get close to a fixed number as $x$ →$0$
