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Lim X → 0 X Cos X + 2 Sin X X 2 + Tan X - Mathematics

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प्रश्न

\[\lim_{x \to 0} \frac{x \cos x + 2 \sin x}{x^2 + \tan x}\] 

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उत्तर

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

Dividing the numerator and the denominator by x, we get:

\[\lim_{x \to 0} \left[ \frac{\cos x + \frac{2\sin x}{x}}{x + \frac{\tan x}{x}} \right]\]
\[ = \frac{\cos0 + 2 \times 1}{0 + 1} \left[ \because \lim_{x \to 0} \left( \frac{\sin x}{x} \right) = 1, \lim_{x \to 0} \left( \frac{\tan x}{x} \right) = 1 \right]\]
\[ = \frac{3}{1}\]

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अध्याय 29: Limits - Exercise 29.7 [पृष्ठ ५०]

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आरडी शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.7 | Q 19 | पृष्ठ ५०

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