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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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