Question

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

Answer 1

We have,
$\left[x\right]=\{\begin{array}{cc}0,& 0\le x<1\\ -1,& -1\le x<0\end{array}$
$\therefore f\left(x\right)=\{\begin{array}{cc}\frac{\sin (-1)}{-1},& -1\le x<0\\ 0,& 0\le x<1\end{array}$
$f\left(x\right)=\{\begin{array}{cc}sin1,& -1\le x<0\\ 0,& 0\le x<1\end{array}$
Now,
${lim}_{x\to 0^{-}}f\left(x\right)={lim}_{x\to 0}sin1=sin1$
${lim}_{x\to 0^{+}}f\left(x\right)={lim}_{x\to 0}0=0$
clearly, ${lim}_{x\to 0^{-}}f\left(x\right)\ne {lim}_{x\to 0^{+}}f\left(x\right)$
thus, ${lim}_{x\to 0}f\left(x\right)$ Does not exist