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Q. The value of $ \displaystyle\lim _{x \rightarrow 0}\left(\left[\frac{100 x}{\sin x}\right]+\left[\frac{99 \sin x}{x}\right]\right)$, where [.] represents greatest integer function, is

Limits and Derivatives

Solution:

We know that $ \displaystyle\lim _{x \rightarrow 0} \frac{\sin x}{x} \rightarrow 1^{-}$
$ \displaystyle\lim _{x \rightarrow 0} \frac{x}{\sin x} \rightarrow 1^{+}$
$\therefore \displaystyle\lim _{x \rightarrow 0}\left[100 \frac{x}{\sin x}\right]+ \displaystyle\lim _{x \rightarrow 0}\left[99 \frac{\sin x}{x}\right]$
$=100+98=198 .$