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प्रश्न
\[\lim_{x \to 0} \frac{\sqrt{1 + x^2} - \sqrt{1 + x}}{\sqrt{1 + x^3} - \sqrt{1 + x}}\]
उत्तर
\[\lim_{x \to 0} \left[ \frac{\sqrt{1 - x^2} - \sqrt{1 + x}}{\sqrt{1 + x^3} - \sqrt{1 + x}} \right]\] It is of the form \[\frac{0}{0}\]
Rationalising the numerator and the denominator:
\[\lim_{x \to 0} \left[ \frac{\left( \sqrt{1 + x^2} - \sqrt{1 + x} \right)}{1} \times \frac{\left( \sqrt{1 + x^2} + \sqrt{1 + x} \right)}{\left( \sqrt{1 + x^2} + \sqrt{1 + x} \right)} \times \frac{1}{\left( \sqrt{1 + x^3} - \sqrt{1 + x} \right)} \times \frac{\left( \sqrt{1 + x^3} - \sqrt{1 + x} \right)}{\left( \sqrt{1 + x^3} + \sqrt{1 + x} \right)} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\left[ \left( 1 + x^2 \right) - \left( 1 + x \right) \right]}{\left[ \left( 1 + x^3 \right) - \left( 1 + x \right) \right]} \times \frac{\left( \sqrt{1 + x^3} + \sqrt{1 + x} \right)}{\left( \sqrt{1 + x^2} + \sqrt{1 + x} \right)} \right]\]
\[ = \lim_{x \to 0} \left[ \left( \frac{x^2 - x}{x^3 - x} \right) \times \frac{\left( \sqrt{1 + x^3} + \sqrt{1 + x} \right)}{\left( \sqrt{1 + x^2} + \sqrt{1 + x} \right)} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{x\left( x - 1 \right)}{x\left( x^2 - 1 \right)} \frac{\left( \sqrt{1 + x^3} + \sqrt{1 + x} \right)}{\left( \sqrt{1 + x^2} + \sqrt{1 + x} \right)} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\left( x - 1 \right) \left( \sqrt{1 + x^3} + \sqrt{1 + x} \right)}{\left( x - 1 \right) \left( x + 1 \right) \left( \sqrt{1 + x^2} + \sqrt{1 + x} \right)} \right]\]
\[ = \frac{\left( \sqrt{1 + 0} + \sqrt{1 + 0} \right)}{\left( 0 + 1 \right) \left( \sqrt{1 + 0} + \sqrt{1 + 0} \right)}\]
\[ = \frac{2}{2}\]
\[ = 1\]
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