Question

If $f\left(x\right)=\{\begin{array}{cc}\frac{\sin \left\{\cos x\right\}}{x-\frac{\pi }{2}},& x\ne \frac{\pi }{2}\\ 1,& x=\frac{\pi }{2}\end{array}$, where {.} represents the fractional part function,

then $\underset{x\to \pi /2}{lim}f\left(x\right)$ is?

• A. $-1$
• B. $1$
• C. Does not exist
• D. $4$

Answer 1

The correct option is C Does not exist

We have,
$LHL=\underset{x\to \pi /2^{-}}{lim}f\left(x\right)=\underset{x\to \pi /2^{-}}{lim}\frac{sin\left\{cosx\right\}}{x-\pi /2}$
$=\underset{h\to 0}{lim}\frac{sin\left\{cos(\frac{\pi }{2}-h)\right\}}{\pi /2-h-\pi /2}=\underset{h\to 0}{lim}\frac{\sin \left\{\sin h\right\}}{-h}$
$=\underset{h\to 0}{lim}\frac{\sin (\sin h-[\sin h\left]\right)}{-h}$
$={lim}_{h\to 0}\frac{\sin \left(\sin h\right)}{-h}$
$={lim}_{h\to 0}\frac{\sin \left(\sin h\right)}{\sin h}\times \frac{\sin h}{-h}=-1$
$\Rightarrow \underset{x\to \pi /2^{-}}{lim}f\left(x\right)=-1$
$RHL=\underset{x\to \pi /2^{+}}{lim}f\left(x\right)=\underset{x\to \pi /2^{+}}{lim}\frac{sin\left\{cosx\right\}}{x-\pi /2}$
$=\underset{h\to 0}{lim}\frac{sin\left\{cos\right(\pi /2+h\left)\right\}}{\frac{\pi }{2}+h-\frac{\pi }{2}}$
$=\underset{h\to 0}{lim}\frac{sin\{-sinh\}}{h}=\underset{h\to 0}{lim}\frac{sin(1-sinh)}{h}\to \infty$

Hence, $\underset{x\to \pi /2}{lim}f\left(x\right)$ does not exist.