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Lim X → 0 √ 1 + 3 X − √ 1 − 3 X X - Mathematics

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प्रश्न

\[\lim_{x \to 0} \frac{\sqrt{1 + 3x} - \sqrt{1 - 3x}}{x}\]

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उत्तर

\[\lim_{x \to 0} \left[ \frac{\sqrt{1 + 3x} - \sqrt{1 - 3x}}{x} \right]\] 

It is of the form \[\frac{0}{0}\] 

Rationalising the numerator: 

\[\lim_{x \to 0} \left[ \frac{\left( \sqrt{1 + 3x} - \sqrt{1 - 3x} \right) \left( \sqrt{1 + 3x} + \sqrt{1 - 3x} \right)}{x \left( \sqrt{1 + 3x} + \sqrt{1 - 3x} \right)} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\left( 1 + 3x \right) - \left( 1 - 3x \right)}{x\left( \sqrt{1 + 3x} + \sqrt{1 - 3x} \right)} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{6x}{x \left( \sqrt{1 + 3x} + \sqrt{1 - 3x} \right)} \right]\]
\[ = \frac{6}{\sqrt{1} + \sqrt{1}}\]
\[ = \frac{6}{2}\]
\[ = 3\]

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अध्याय 29: Limits - Exercise 29.4 [पृष्ठ २९]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.4 | Q 25 | पृष्ठ २९

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