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प्रश्न
\[\lim_{x \to 1} \frac{1 - x^2}{\sin \pi x}\]
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उत्तर
\[ \lim_{h \to 0} \frac{1 - \left( 1 - h \right)^2}{\sin \pi\left( 1 - h \right)}\]
\[ \lim_{h \to 0} \frac{2h - h^2}{\sin \pi h}\]
\[ = \lim_{h \to 0} \frac{h\left( 2 - h \right)}{\sin \pi h}\]
\[ = \lim_{h \to 0} \frac{\left( 2 - h \right)}{\pi \times \frac{\sin \pi h}{\pi h}}\]
\[ \Rightarrow \frac{2 - 0}{\pi} = \frac{2}{\pi}\]
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