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प्रश्न
\[\lim_{x \to 0} \frac{e^x - x - 1}{2}\]
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उत्तर
\[\lim_{x \to 0} \left[ \frac{e^x - x - 1}{2} \right]\]
\[ e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + . . . . . \infty \]
\[ = \lim_{x \to 0} \left[ \frac{\left( 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} . . . . \infty \right) - x - 1}{2} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\frac{x^2}{2!} + \frac{x^3}{3!} + . . . . \infty}{2} \right]\]
\[ = 0\]
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