Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{[x{]}^2+sin[x]}{\left[x\right]}& for\left[x\right]\ne 0\\ 0for\left[x\right]=0& \end{array}$ where [x] denotes the greatest integer function,

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

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

Answer 1

The correct option is D Does not exist

$LHL={lim}_{x\to 0^{-}}f\left(x\right)={lim}_{h\to 0}f(0-h)$
$=\underset{h\to 0}{lim}\frac{[-h{]}^2+sin[-h]}{[-h]}$
$=\underset{h\to 0}{lim}\left(\right[-h]+\frac{\sin [-h]}{[-h]})$
$=-1+\frac{\sin (-1)}{(-1)}=-1+\sin 1$
$RHL={lim}_{x\to 0^{+}}f\left(x\right)={lim}_{h\to 0}f(0+h)$
$={lim}_{h\to 0}\left(\right[h]+\frac{\sin \left[h\right]}{\left[h\right]})$
$=0+\frac{\sin \left(0\right)}{0}=0$
Since, $LHL\ne RHL$
Hence, ${lim}_{x\to 0}f\left(x\right)$ does not exists.