Question

If $f\left(x\right)=\{\begin{array}{c}\frac{sin\left[x\right]}{\left[x\right]},\left[x\right]\ne 0\\ 0,\left[x\right]=0\end{array}where\left[x\right]$denotes the greatest integer less than or equal to x, then

• A. $\underset{x\to 0^{-}}{lim}f\left(x\right)=sin1$
• B. $\underset{x\to 0^{+}}{lim}f\left(x\right)=0$
• C. limit does not exist at $x=0$
• D. limit exist at $x=0$

Answer 1

Answer: (A), (B), (C)

$\underset{x\to 0^{+}}{lim}f\left(x\right)=0$
$\underset{x\to 0^{-}}{lim}f\left(x\right)=sin1$
limit does not exist at $x=0$

$f\left(x\right)=\{\begin{array}{c}\frac{\sin \left[x\right]}{\left[x\right]}\left[x\right]\ne 0\\ 0\left[x\right]=0\end{array}$
$\Rightarrow f\left(x\right)=\{\begin{array}{c}\frac{\sin \left[x\right]}{\left[x\right]}xϵR-(0,1)\\ 00\le x<1\end{array}$
At $x=0$ $RHL=\underset{x\to 0+}{lim}0=0$
and $LHL=\underset{x\to 0-}{lim}\frac{\sin \left[x\right]}{\left[x\right]}=\underset{h\to 0}{lim}\frac{\sin [o-h]}{[o-h]}$
$=\underset{h\to 0}{lim}\frac{\sin (-1)}{-1}=\sin 1$