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