Question

If \[f\left( x \right) = x \sin \left( 1/x \right), x \neq 0,\]  then \[\lim_{x \to 0} f\left( x \right) =\] 

पर्याय

• (a) 1 

• (b) 0

• (c) −1 

• (d) does not exist

Answer 1

(b)  0 

\[\lim_{x \to 0} f\left( x \right) = \lim_{x \to 0} x \sin \frac{1}{x}, x \neq 0\]
\[\text{ LHL at } x = 0: \]
\[ \lim_{x \to 0^-} f\left( x \right) = \lim_{h \to 0} f\left( 0 - h \right)\]
\[ \Rightarrow \lim_{h \to 0} - h \sin \left( \frac{1}{- h} \right)\]
\[ = 0\]
\[\text{ RHL } at x = 0: \]
\[ \lim_{x \to 0^+} f\left( x \right) = \lim_{h \to 0} f\left( 0 + h \right)\]
\[ \Rightarrow \lim_{h \to 0} h \sin \left( \frac{1}{h} \right)\]
\[ = 0\]
\[ \lim_{x \to 0^-} f\left( x \right) = \lim_{x \to 0^+} f\left( x \right)\]
\[\text{ Hence }, \lim_{x \to 0} f\left( x \right) \text{`exists and is equal to 0 } .\]