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