Question

Let $f\left(x\right)=\{\begin{array}{c}\frac{\sin \left[x\right]}{\left[x\right]};\left[x\right]\ne 0\\ 0;\left[x\right]=0\end{array}$, then ${lim}_{x\to 0}f\left(x\right)=$

• A. $0$
• B. $\sin 1$
• C. $2$
• D. does not exist

Answer 1

The correct option is D does not exist

$L={lim}_{x\to 0}\frac{\sin \left[x\right]}{\left[x\right]}$
Left Hand Limit
$={lim}_{h\to 0}f(0-h)={lim}_{h\to 0}\frac{\sin [-h]}{[-h]}=\frac{-\sin 1}{-1}=\sin 1$
Right Hand Limit
$={lim}_{h\to 0}f(0+h)={lim}_{h\to 0}\frac{\sin \left[h\right]}{\left[h\right]}=\frac{\sin 0}{0}=f\left(0\right)=0$
$\because$ Left hand limit $\left(f\right(0^{-}\left)\right)\ne$ Right hand limit $\left(f\right(0^{+}\left)\right)$
$\therefore$ Limit of $f\left(x\right)$ at $x=0$ does not exist.