Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{\tan \left(\right[x]+x)}{\left[x\right]-2x},& x\ne 0\\ 0,& x=0\end{array}$, then $\underset{x\to 0}{lim}f\left(x\right)$ equal to
(where $[.]$ denotes the greatest integer function)

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

Answer 1

The correct option is D does not exist

$L.H.L.=\underset{x\to 0^{-}}{lim}f\left(x\right)$
$=\underset{h\to 0}{lim}f(0-h)$
$=\underset{h\to 0}{lim}\frac{\tan \left(\right[-h]-h)}{[-h]+2h}$
$=\underset{h\to 0}{lim}\frac{\tan (-1-h)}{2h-1}$
$=\underset{h\to 0}{lim}\frac{\tan (1+h)}{1-2h}$
$=\frac{\tan 1}{1}=\tan 1$

$R.H.L=\underset{x\to 0^{+}}{lim}f\left(x\right)$
$=\underset{h\to 0}{lim}f(0+h)$
$=\underset{h\to 0}{lim}\frac{\tan \left(\right[h]+h)}{\left[h\right]-2h}$
$=\underset{h\to 0}{lim}\frac{\tan \left(h\right)}{-2h}$
$=-\frac{1}{2}\times 1=-\frac{1}{2}$
Since $L.H.L\ne R.H.L$
$\therefore \underset{x\to 0}{lim}f\left(x\right)$ does not exist.