मराठी

P Lim X → 0 √ 1 + X − 1 Log ( 1 + X )

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प्रश्न

\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{\log \left( 1 + x \right)}\] 

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उत्तर

\[\lim_{x \to 0} \left[ \frac{\sqrt{1 + x} - 1}{\log \left( 1 + x \right)} \right]\] 

Rationalising the numerator: 

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

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पाठ 29: Limits - Exercise 29.1 [पृष्ठ ७१]

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आर.डी. शर्मा Mathematics [English] Class 11
पाठ 29 Limits
Exercise 29.1 | Q 26 | पृष्ठ ७१

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