Mathematics - Limits Question with Solution | TestHub

LHL :

${\lim }_{x\to \alpha -}\left[\frac{\min (\sin x,x-[x]}{(x-1)}\right]$

When $1<x<\alpha$

$\{x\}=x-1<\sin x$

$\min \{\sin x,x-1\}=x-1$

Required limit $={\lim }_{x\to \infty -}\left[\frac{x-1}{x-1}\right]=1$

$\begin{array}{rlr}& x\to \alpha +& \sin x<x-1\end{array}$

RHL : ${\lim }_{x\to \alpha +}\left[\frac{\sin x}{x-1}\right]=0$

$\frac{\sin x}{x-1}<1$

Hence $LHL\ne RHL$ $\left[\frac{\sin x}{x-1}\right]=0$

Limit does not exist