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प्रश्न
\[\lim_{x \to 0} \frac{x\left( 2^x - 1 \right)}{1 - \cos x}\]
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उत्तर
\[\lim_{x \to 0} \left[ \frac{x\left( 2^x - 1 \right)}{1 - \cos x} \right]\]
Dividing the numerator and the denominator by x2:
\[= \lim_{x \to 0} \left[ \left( \frac{2^x - 1}{x} \right) \times \frac{x^2}{1 - \cos x} \right]\]
\[ = \lim_{x \to 0} \left[ \left( \frac{2^x - 1}{x} \right) \times \frac{x^2}{2 \sin^2 \frac{x}{2}} \right]\]
\[ = \lim_{x \to 0} \left[ \left( \frac{2^x - 1}{x} \right) \times \frac{\frac{x^2}{4} \times 4}{2 \sin^2 \left( \frac{x}{2} \right)} \right]\]
\[ = \left( \log 2 \right) \times \frac{4}{2} \times \frac{1}{1^2}\]
\[ = 2 \log 2\]
\[ = \log \left( 2 \right)^2 \]
\[ = \log 4\]
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