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प्रश्न
\[\lim_{x \to \pi} \frac{\sin x}{\pi - x}\]
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उत्तर
\[\lim_{x \to \pi} \frac{\sin x}{\pi - x}\]
\[ = \lim_{h \to 0} \frac{\sin \left( \pi - h \right)}{\pi - \left( \pi - h \right)} \left[ \because \lim_{x \to a} f\left( x \right) = \lim_{h \to 0} f\left( a - h \right) \right]\]
\[ = \lim_{h \to 0} \frac{\sin h}{h} \left[ \because \sin \left( \pi - 0 \right) = \sin 0 \right]\]
\[ \Rightarrow 1\]
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