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