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Lim X → 0 Log ( 3 + X ) − Log ( 3 − X ) X

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प्रश्न

\[\lim_{x \to 0} \frac{\log \left( 3 + x \right) - \log \left( 3 - x \right)}{x}\] 

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उत्तर

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

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अध्याय 29: Limits - Exercise 29.1 [पृष्ठ ७१]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.1 | Q 23 | पृष्ठ ७१

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