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प्रश्न
\[\lim_{x \to 0^+} \left\{ 1 + \tan^2 \sqrt{x} \right\}^{1/2x}\]
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उत्तर
\[\lim_{x \to 0^+} \left[ 1 + \tan^2 \sqrt{x} \right] {}^\frac{1}{2x} \]
\[\text{ Using the theorem given below }: \]
\[\text{ If } \lim_{x \to a} f\left( x \right) = \lim_{x \to a} g\left( x \right) = 0 \text{ such that } \lim_{x \to a} \frac{f\left( x \right)}{g\left( x \right)} \text{ exists, then } \lim_{x \to a} \left[ 1 + f\left( x \right) \right]^\frac{1}{g\left( x \right)} = e^\lim_{x \to a} \frac{f\left( x \right)}{g\left( x \right)} . \]
\[\text{ Here }: \]
\[ f\left( x \right) = \tan^2 \sqrt{x}\]
\[ g\left( x \right) = 2x\]
\[ \Rightarrow e^\lim_{x \to 0^+} \left( \frac{\tan^2 \sqrt{x}}{2x} \right) \]
\[ = e^\lim_{x \to 0^+} \left( \frac{\tan \sqrt{x}}{\sqrt{x}} \right) \times \left( \frac{\tan \sqrt{x}}{\sqrt{x}} \right) \times \frac{1}{2} \]
\[ = e^{1 \times 1 \times \frac{1}{2}} \]
\[ = \sqrt{e}\]
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