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Lim X → 3 ( X 2 − 9 ) [ 1 X + 3 + 1 X − 3 ] - Mathematics

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प्रश्न

\[\lim_{x \to 3} \left( x^2 - 9 \right) \left[ \frac{1}{x + 3} + \frac{1}{x - 3} \right]\] 

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उत्तर

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

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अध्याय 29: Limits - Exercise 29.3 [पृष्ठ २३]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.3 | Q 24 | पृष्ठ २३

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