Question

If \[f\left( x \right) = \left\{ \begin{array}{l}x \sin \frac{1}{x}, & x \neq 0 \\ 0, & x = 0\end{array}, \right.\] then \[\lim_{x \to 0} f\left( x \right)\]  equals 

पर्याय

•  1 

•  0 

•  −1 

•  none of these 

Answer 1

 0 

\[f\left( x \right) = \binom{x\sin\left( \frac{1}{x} \right), x \neq 0}{0, x = 0}\] 

LHL: 

\[\lim_{x \to 0^-} f\left( x \right)\]
\[ = \lim_{x \to 0^-} \left[ x\sin\left( \frac{1}{x} \right) \right]\]

Let x = 0 – h, where h → 0. 

\[= \lim_{h \to 0} \left[ \left( - h \right) \times \sin\left( - \frac{1}{h} \right) \right]\] 

= 0 × The oscillating number between –1 and 1
= 0
RHL 

\[\lim_{x \to 0^+} f\left( x \right)\] 

Let x = 0 + h, where h → 0.

\[= \lim_{h \to 0} \left[ h \times \sin\left( \frac{1}{h} \right) \right]\] 

= 0 × The oscillating number between –1 and 1
= 0
LHL = RHL = 0 

\[\therefore \lim_{x \to 0} f\left( x \right) = 0\]