मराठी

Lim X → 0 √ 1 + X − 1 X

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प्रश्न

\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}\] 

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उत्तर

\[\lim_{x \to 0} \left[ \frac{\sqrt{1 + x} - 1}{x} \right]\] It is of the form \[\frac{0}{0}\]

Rationalising the numerator: 

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

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

=  \[\frac{1}{\sqrt{1 + 0} + 1}\] 

=  \[\frac{1}{2}\]

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पाठ 29: Limits - Exercise 29.4 [पृष्ठ २८]

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आर.डी. शर्मा Mathematics [English] Class 11
पाठ 29 Limits
Exercise 29.4 | Q 12 | पृष्ठ २८

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