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Limx→0xtan⁡x1−cos⁡x - Mathematics

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प्रश्न

\[\lim_{x \to 0} \frac{x \tan x}{1 - \cos x}\]

बेरीज
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उत्तर

\[\lim_{x \to 0} \left[ \frac{x \tan x}{1 - \cos x} \right]\]

\[ {\text{ Dividing the numerator and the denominator by } x}^2:\]

\[ \lim_{x \to 0} \left[ \frac{\frac{x \tan x}{x^2}}{\frac{1 - \cos x}{x^2}} \right]\]

\[ = \lim_{x \to 0} \left[ \frac{\frac{\tan x}{x}}{\frac{2 \sin^2 \frac{x}{2}}{\frac{x}{2} \times \frac{x}{2} \times 4}} \right]\]

\[ = \frac{4}{2}\]

\[ = 2\]

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पाठ 29: Limits - Exercise 29.7 [पृष्ठ ५०]

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आरडी शर्मा Mathematics [English] Class 11
पाठ 29 Limits
Exercise 29.7 | Q 35 | पृष्ठ ५०

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