Question

$\underset{x\to 0}{lim}\left[\right(1-e^x\left)\frac{\sin x}{|x|}\right]$ is (where [.] represents the greatest integer function)

• A. $-1$
• B. $1$
• C. $0$
• D. Does not exist

Answer 1

The correct option is A $-1$

Given,

$\underset{x\to 0}{lim}\left[\right(1-e^x\left)\frac{\sin x}{|x|}\right]$
Now, at $x\to 0^{-}$

LHL $=\underset{x\to 0^{-}}{lim}\left[\right(1-e^x\left)\frac{\sin x}{|x|}\right]=\underset{h\to 0}{lim}\left[\right(1-e^{-h}\left)\frac{\sin (-h)}{h}\right]$

$=\underset{h\to 0}{lim}\left[\frac{-h(e^{-h}-1)}{-h}\frac{\sin h}{h}\right]=\underset{h\to 0}{lim}[-h]=-1$
At $x\to 0^{+}$
RHL $=\underset{x\to 0^{+}}{lim}\left[\right(1-e^x\left)\frac{\sin x}{|x|}\right]=\underset{h\to 0}{lim}\left[\right(1-e^h\left)\frac{\sin h}{h}\right]$

$=\underset{h\to 0}{lim}\left[\frac{-h(e^h-1)}{h}\frac{\sin h}{h}\right]=\underset{h\to 0}{lim}[-1]=-1$

Hence, $\underset{x\to 0}{lim}\left[\right(1-e^x\left)\frac{\sin x}{|x|}\right]=-1$