Question

Let $\left\{x\right\}$ denote the fractional part of $x$. Then $\underset{x\to 0}{lim}\frac{\left\{x\right\}}{\tan \left\{x\right\}}$ is equal to

• A. $-1$
• B. $0$
• C. $1$
• D. Does not exist

Answer 1

The correct option is D Does not exist

$\left\{x\right\}=x-\left[x\right]$ and $0\le \left\{x\right\}<1$
${lim}_{x\to 0^{+}}\frac{\left\{x\right\}}{\tan \left\{x\right\}}={lim}_{x\to 0^{+}}\frac{x-\left[x\right]}{\tan (x-\left[x\right])}$
${lim}_{x\to 0^{+}}\frac{x}{\tan x}={lim}_{x\to 0^{+}}\frac{1}{{\sec }^{2}x}=1$
[ use L-Hospital rule]
${lim}_{x\to 0^{-}}\frac{\left\{x\right\}}{\tan \left\{x\right\}}={lim}_{x\to 0^{-}}\frac{x-\left[x\right]}{\tan (x-\left[x\right])}$
$={lim}_{x\to 0^{-}}\frac{x-[-1]}{\tan (x-[-1])}$
$={lim}_{x\to 0^{-}}\frac{x+1}{\tan (x+1)}$
$=\frac{1}{\tan 1}$
$\because {lim}_{x\to 0^{+}}\frac{\left\{x\right\}}{\tan \left\{x\right\}}\ne {lim}_{x\to 0^{-}}\frac{\left\{x\right\}}{\tan \left\{x\right\}}$
So, ${lim}_{x\to 0}\frac{\left\{x\right\}}{\tan \left\{x\right\}}$ does not exist.