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Lim X → 0 Sin 2 X E X − 1 - Mathematics

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प्रश्न

\[\lim_{x \to 0} \frac{\sin 2x}{e^x - 1}\] 

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उत्तर

\[\lim_{x \to 0} \left[ \frac{\sin \left( 2x \right)}{e^x - 1} \right]\]

Dividing the numerator and the denominator by x

\[= \lim_{x \to 0} \left[ \frac{\sin 2x}{x \times \frac{e^x - 1}{x}} \right]\]

\[ = \lim_{x \to 0} \left[ \frac{\sin 2x}{2x} \times \frac{2}{\left( \frac{e^x - 1}{x} \right)} \right]\]

\[ = 1 \times \frac{2}{1}\]

\[ = 2\]

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पाठ 29: Limits - Exercise 29.1 [पृष्ठ ७१]

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आरडी शर्मा Mathematics [English] Class 11
पाठ 29 Limits
Exercise 29.1 | Q 16 | पृष्ठ ७१

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