Question

Let $\left\{x\right\}$ denotes the fraction 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. $noneofthese$

Answer 1

The correct option is D $noneofthese$

We have, $\underset{x\to 0^{-}}{lim}\frac{\left\{x\right\}}{\tan \left\{x\right\}}$

$=\underset{x\to 0^{-}}{lim}\frac{x-\left[x\right]}{\tan (x-[x\left]\right)}=\underset{x\to 0^{-}}{lim}\frac{x+1}{\tan (x+1)}=\frac{1}{\tan 1}=cot1$ and,

$\underset{x\to 0^{+}}{lim}\frac{\left\{x\right\}}{\tan \left\{x\right\}}=\underset{x\to 0^{+}}{lim}\frac{x-\left[x\right]}{\tan (x-[x\left]\right)}=\underset{x\to 0^{+}}{lim}\frac{x}{\tan x}=1$

Clearly, $\underset{x\to 0^{-}}{lim}\frac{\left\{x\right\}}{\tan \left\{x\right\}}\ne \underset{x\to 0^{+}}{lim}\frac{\left\{x\right\}}{\tan \left\{x\right\}}$

So, $\underset{x\to 0}{lim}\frac{\left\{x\right\}}{\tan \left\{x\right\}}$ does not exist.