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प्रश्न
\[\lim_{x \to 0} \left\{ \frac{e^x + e^{- x} - 2}{x^2} \right\}^{1/ x^2}\]
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उत्तर
\[\lim_{x \to 0} \left[ \frac{e^x + e^{- x} - 2}{x^2} \right]^\left( \frac{1}{x^2} \right) \]
\[ = \lim_{x \to 0} \left[ 1 + \frac{e^x + e^{- x} - 2}{x^2} - 1 \right]^\left( \frac{1}{x^2} \right) \]
\[ = e {}^\lim_{x \to 0} \left( \frac{e^x + e^{- x} - 2}{x^2} - 1 \right) \times \left( \frac{1}{x^2} \right) \]
\[ \because e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + . . . . . . \propto \]
\[ e^{- x} = 1 - \frac{x}{1!} + \frac{x^2}{2!} - \frac{x^3}{3!} + . . . . . . \propto \]
\[ \Rightarrow e^x + e^{- x} = 2 + \frac{2 x^2}{2!} + \frac{2 x^4}{4!} + . . . . . \propto \]
\[ = e^\lim_{x \to 0} \left( \frac{2 + \frac{2 x^2}{2!} + \frac{2 x^4}{4!} . . . \propto - 2}{x^2} - 1 \right) \times \left( \frac{1}{x^2} \right) \]
\[ = e^\lim_{x \to 0} \left( \frac{\frac{2 x^2}{2!} + \frac{2 x^4}{4!} + . . . . . . \propto}{x^4} - \frac{1}{x^2} \right) \]
\[ = e^\lim_{x \to 0} \left( \frac{x^2 + \frac{x^4}{12} + . . . . . \propto - x^2}{x^4} \right) \]
\[ = e^\frac{1}{12}\]
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