Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{\sin (1+[x\left]\right)}{\left[x\right]}& \text{for}\left[x\right]\ne 0\\ 0& \text{for}\left[x\right]=0\end{array}$
where $\left[x\right]$ denotes the greatest integer not exceeding $x$ then $\underset{x\to 0^{-}}{lim}f\left(x\right)=$

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is B $0$

$\underset{x\to 0^{-}}{lim}f\left(x\right)=\underset{x\to 0^{-}}{lim}\frac{\sin (1+[x\left]\right)}{\left[x\right]}$
$=\frac{\sin (1+[0^{-}\left]\right)}{\left[0^{-}\right]}=\frac{\sin (1-1)}{-1}=-\frac{\sin 0}{1}=0$
Since $\left[0^{-}\right]=-1$