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प्रश्न
\[\lim_{x \to \pi} \frac{\sin x}{x - \pi} .\]
उत्तर
\[\lim_{x \to \pi} \left( \frac{\sin x}{x - \pi} \right)\]
\[\text{ LHL }: \]
\[ \lim_{x \to \pi^-} \left( \frac{\sin x}{x - \pi} \right)\]
\[\text{ If } x = \pi - h, \text{ then } h \to 0 . \]
\[ = \lim_{h \to 0} \left( \frac{\sin \left( \pi - h \right)}{\pi - h - \pi} \right)\]
\[ = \lim_{h \to 0} \left( \frac{\sin h}{- h} \right)\]
\[ = - 1\]
\[\text{ RHL }: \]
\[ \lim_{x \to \pi^+} \left( \frac{\sin x}{x - \pi} \right)\]
\[If x = \pi + h, \text{ then } h \to 0 . \]
\[ = \lim_{h \to 0} \left( \frac{\sin \left( \pi + h \right)}{\pi + h - \pi} \right)\]
\[ = \lim_{h \to 0} \left( \frac{- \sin h}{h} \right)\]
\[ = - 1\]
\[ \therefore \lim_{x \to \pi} \left( \frac{\sin x}{x - \pi} \right) = - 1\]
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