Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{ta{n}^{-1}(x+[x\left]\right)}{\left[x\right]-2x},& \left[x\right]\ne 0\\ 0,& \left[x\right]=0\end{array}$ where [x] denotes the greatest integer less than or equal to $x$

then $\underset{x\to 0}{lim}f\left(x\right)$ is?

• A. $-\frac{1}{2}$
• B. $1$
• C. $\frac{\pi }{4}$
• D. Does not exist

Answer 1

Answer: (D)

Does not exist

We have,
$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(0-h)=\underset{h\to 0}{lim}f(-h)$

$\Rightarrow \underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}\frac{ta{n}^{-1}(-h+[-h\left]\right)}{[-h]+2h}=\underset{h\to 0}{lim}\frac{ta{n}^{-1}(-1-h)}{-1+2h}$

$\Rightarrow \underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{h\to 0}{lim}\frac{ta{n}^{-1}(1+h)}{1-2h}=ta{n}^{-1}1=\frac{\pi }{4}$ and,

$\underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}f(0+h)=\underset{h\to 0}{lim}f\left(h\right)$

$\Rightarrow \underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}\frac{ta{n}^{-1}(h+[h\left]\right)}{\left[h\right]-2h}$

$\Rightarrow \underset{x\to 0^{+}}{lim}f\left(x\right)=\underset{h\to 0}{lim}\frac{ta{n}^{-1}h}{-2h}=-\frac{1}{2}$ $[\because [h]=0]$

Hence, $\underset{x\to 0}{lim}f\left(x\right)$ does not exist.