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Lim X → 0 9 X − 2 . 6 X + 4 X X 2 - Mathematics

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प्रश्न

\[\lim_{x \to 0} \frac{9^x - 2 . 6^x + 4^x}{x^2}\] 

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उत्तर

\[\lim_{x \to 0} \left[ \frac{9^x - 2 . 6^x + 4^x}{x^2} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\left( 3^x \right)^2 - 2 \cdot 3^x \cdot 2^x + \left( 2^x \right)^2}{x^2} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\left( 3^x - 2^x \right)^2}{x^2} \right]\]
\[ = \lim_{x \to 0} \left[ \left( \frac{3^x - 2^x}{2^x} \right)^2 \times \frac{\left( 2^x \right)^2}{x^2} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\left( \frac{3}{2} \right)^x - 1}{x} \right]^2 \times 2^{2x} \]
\[ = \left[ \log \left( \frac{3}{2} \right) \right]^2 \times 2^0 \]
\[ = \left[ \log \left( \frac{3}{2} \right) \right]^2\]

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अध्याय 29: Limits - Exercise 29.1 [पृष्ठ ७१]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.1 | Q 6 | पृष्ठ ७१

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