मराठी

Lim X → 0 √ 2 − √ 1 + Cos X X 2 - Mathematics

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प्रश्न

\[\lim_{x \to 0} \frac{\sqrt{2} - \sqrt{1 + \cos x}}{x^2}\] 

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उत्तर

\[\lim_{x \to 0} \left[ \frac{\sqrt{2} - \sqrt{1 + \cos x}}{x^2} \right]\]
\[\text{ Rationalising the numerator }: \]
\[ \lim_{x \to 0} \left[ \frac{\left( \sqrt{2} - \sqrt{1 + \cos x} \right) \left( \sqrt{2} + \sqrt{1 + \cos x} \right)}{x^2 \times \left( \sqrt{2} + \sqrt{1 + \cos x} \right)} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{2 - \left( 1 + \cos x \right)}{x^2 \left[ \sqrt{2} + \sqrt{1 + \cos x} \right]} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{1 - \cos x}{x^2 \left[ \sqrt{2} + \sqrt{1 + \cos x} \right]} \right]\]
\[ = \lim_{x \to 0} \left[ \frac{2 \sin^2 \left( \frac{x}{2} \right)}{x^2 \left\{ \sqrt{2} + \sqrt{1 + \cos x} \right\}} \right]\]
\[ = 2 \lim_{x \to 0} \left[ \frac{\sin^2 \left( \frac{x}{2} \right)}{4 \times \left( \frac{x}{2} \right)^2} \times \frac{1}{\sqrt{2} + \sqrt{1 + \cos x}} \right]\]
\[ = 2 \times \frac{1}{4} \times \frac{1}{\sqrt{2} + \sqrt{1 + \cos 0}}\]
\[ = \frac{2}{4\left( \sqrt{2} + \sqrt{2} \right)}\]
\[ = \frac{1}{4\sqrt{2}}\]

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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 34 | पृष्ठ ५०

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