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प्रश्न
\[\lim_{x \to 0} \frac{e^x - 1}{\sqrt{1 - \cos x}}\]
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उत्तर
\[\lim_{x \to 0} \left[ \frac{e^x - 1}{\sqrt{1 - \cos x}} \right]\]
\[\text{ Rationalising the denominator, we get }: \]
\[ = \lim_{x \to 0} \left[ \frac{\left( e^x - 1 \right)}{\sqrt{1 - \cos x}} \times \frac{\sqrt{1 + \cos x}}{\sqrt{1 + \cos x}} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\left( e^x - 1 \right) \left( \sqrt{1 + \cos x} \right)}{\sqrt{1 - \cos^2 x}} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{\left( e^x - 1 \right) \sqrt{1 + \cos x}}{\left| \sin x \right|} \right]\]
Dividing numerator and the denominator by x, we get:
\[\text{ Left hand limit }: \]
\[ \lim_{x \to 0^-} \left[ \left( \frac{e^x - 1}{x} \right) \times \frac{\sqrt{1 + \cos x}}{\left( \frac{\left| \sin x \right|}{x} \right)} \right]\]
\[ = \lim_{x \to 0^-} \left[ \left( \frac{e^x - 1}{x} \right) \times \frac{\sqrt{1 + \cos x}}{\left( \frac{- \sin x}{x} \right)} \right]\]
\[ = - \frac{1 \times \sqrt{2}}{1}\]
\[ = - \sqrt{2}\]
\[\text{ Right hand limit }: \]
\[ \lim_{x \to 0^+} \left[ \left( \frac{e^x - 1}{x} \right) \times \frac{\sqrt{1 + \cos x}}{\frac{\left| \sin x \right|}{x}} \right]\]
\[ = \lim_{x \to 0^+} \left[ \left( \frac{e^x - 1}{x} \right) \times \frac{\sqrt{1 + \cos x}}{\frac{\sin x}{x}} \right]\]
\[ = 1 \times \frac{\sqrt{2}}{1}\]
\[ = \sqrt{2}\]
Left hand limit ≠ Right hand limit
Thus, limit does not exist.
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