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प्रश्न
\[\lim_{x \to 0} \frac{8^x - 4^x - 2^x + 1}{x^2}\]
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उत्तर
\[\lim_{x \to 0} \left[ \frac{8^x - 4^x - 2^x + 1}{x^2} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\left( 2^x \right)^3 - \left( 2^x \right)^2 - 2^x + 1}{x^2} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\left( 2^x \right)^2 \left( 2^x - 1 \right) - 1\left( 2^x - 1 \right)}{x^2} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\left( 2^{2x} - 1 \right) \left( 2^x - 1 \right)}{x^2} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{2\left( 2^{2x} - 1 \right)}{2x} \times \left( \frac{2^x - 1}{x} \right) \right]\]
\[ = 2 \log 2 \times \log 2\]
\[ = \log \left( 2 \right)^2 \times \log 2\]
\[ = \left( \log 4 \right) \times \left( \log 2 \right)\]
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