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प्रश्न
\[\lim_{x \to 0} \frac{8}{x^8}\left\{ 1 - \cos \frac{x^2}{2} - \cos \frac{x^2}{4} + \cos \frac{x^2}{2} \cos \frac{x^2}{4} \right\}\] is equal to
पर्याय
\[\frac{1}{16}\]
\[- \frac{1}{16}\]
\[\frac{1}{32}\]
\[- \frac{1}{32}\]
उत्तर
\[\frac{1}{32}\]
\[\lim_{x \to 0} \frac{8}{x^8} \left[ 1 - \cos \frac{x^2}{2} - \cos\frac{x^2}{4} + \cos\frac{x^2}{2}\cos \frac{x^2}{4} \right]\]
\[ = \lim_{x \to 0} \frac{8}{x^8} \left[ \left( 1 - \cos \frac{x^2}{4} \right) - \cos \frac{x^2}{2}\left( 1 - \cos\frac{x^2}{4} \right) \right]\]
\[ = \lim_{x \to 0} \frac{8}{x^8} \left[ \left( 1 - \cos\frac{x^2}{4} \right) \left( 1 - \cos\frac{x^2}{2} \right) \right]\]
\[ = \lim_{x \to 0} \frac{8}{x^8} \left[ \left( 2 \sin^2 \frac{x^2}{8} \right) \left( 2 \sin^2 \frac{x^2}{4} \right) \right]\]
\[ = \lim_{x \to 0} 4 \times 8 \frac{\left( \sin^2 \frac{x^2}{8} \right)}{\left( 64 \times \frac{x^4}{64} \right)} \frac{\left( \sin^2 \frac{x^2}{4} \right)}{16\left( \frac{x^4}{16} \right)}\]
\[ = \frac{32}{64 \times 16}\]
\[ = \frac{1}{32}\]
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