Question

Let $f\left(x\right)=\frac{x+|x|(1+x)}{x}\sin \left(\frac{1}{x}\right),x\ne 0$
Write $L=\underset{x\to 0^{-}}{lim}f\left(x\right)$ and $R=\underset{x\to 0^{+}}{lim}f\left(x\right).$ Then

• A. $L$ exists but $R$ does not exist
• B. $L$ does not exist but $R$ exists
• C. Both $L$ and $R$ exist
• D. Neither $L$ nor $R$ exists.

Answer 1

The correct option is A $L$ exists but $R$ does not exist

$L=\underset{x\to 0^{-}}{lim}f\left(x\right)$
$=\underset{h\to 0}{lim}\frac{-h+|-h|(1-h)}{-h}\sin \left(\frac{1}{-h}\right)$
$=\underset{h\to 0}{lim}\frac{-h+h(1-h)}{h}\sin \left(\frac{1}{h}\right)$
$=\underset{h\to 0}{lim}\frac{h(1-h-1)}{h}\sin \left(\frac{1}{h}\right)$
$=\underset{h\to 0}{lim}-h\sin \left(\frac{1}{h}\right)=0$

$R=\underset{x\to 0^{+}}{lim}f\left(x\right)$
$=\underset{h\to 0}{lim}\frac{h+|h|(1+h)}{h}\sin \left(\frac{1}{h}\right)$
$=\underset{h\to 0}{lim}\frac{h+h(1+h)}{h}\sin \left(\frac{1}{h}\right)$
$=\underset{h\to 0}{lim}\frac{h(1+1+h)}{h}\sin \left(\frac{1}{h}\right)$
$=\underset{h\to 0}{lim}(2+h)\sin \left(\frac{1}{h}\right)$
$=\underset{h\to 0}{lim}2\sin \left(\frac{1}{h}\right)+\underset{h\to 0}{lim}h\sin \left(\frac{1}{h}\right)$
The first limit does not exist and hence, $R$ does not exist.