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lim x → 0 7 x cos x − 3 sin x 4 x + tan x

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प्रश्न

\[\lim_{x \to 0} \frac{7x \cos x - 3 \sin x}{4x + \tan x}\] 

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उत्तर

\[\lim_{x \to 0} \left[ \frac{7x \cos x - 3 \sin x}{4x + \tan x} \right]\]  It is of the form \[\left( \frac{0}{0} \right)\] 

Dividing the numerator and the denominator by x

\[\Rightarrow \lim_{x \to 0} \frac{7\cos x - 3\left( \frac{\sin x}{x} \right)}{4 + \left( \frac{\tan x}{x} \right)}\]
\[ \Rightarrow \frac{7 \lim_{x \to 0} \left( \cos x \right) - 3 \lim_{x \to 0} \left( \frac{\sin x}{x} \right)}{4 + \lim_{x \to 0} \left( \frac{\tan x}{x} \right)}\]
\[ \Rightarrow \frac{7 . 1 - 3 . 1}{4 + 1}\]
\[ \Rightarrow \frac{4}{5}\]

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अध्याय 29: Limits - Exercise 29.7 [पृष्ठ ५०]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.7 | Q 10 | पृष्ठ ५०

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