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Lim X → 0 5 X Cos X + 3 Sin X 3 X 2 + Tan X - Mathematics

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प्रश्न

\[\lim_{x \to 0} \frac{5 x \cos x + 3 \sin x}{3 x^2 + \tan x}\] 

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उत्तर

\[\lim_{x \to 0} \left[ \frac{5 x \cos x + 3 \sin x}{3 x^2 + \tan x} \right]\] Dividing the numerator and the denominator by x: 

\[\lim_{x \to 0} \left[ \frac{5\cos x + 3\left( \frac{\sin x}{x} \right)}{3x + \left( \frac{\tan x}{x} \right)} \right]\]
\[ = \left[ \frac{5\cos0 + 3}{3 \times 0 + 1} \right] \left[ \because \lim_{x \to 0} \frac{\sin x}{x} = 1, \lim_{x \to 0} \frac{\tan x}{x} = 1 \right]\]
\[ = \frac{5 + 3}{0 + 1}\]
\[ = 8\] 

 

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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 21 | पृष्ठ ५०

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