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

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

Hence $LHL\ne RHL$

$$\left[\frac{\sin x}{x-1}\right]=0$$

Limit does not exist