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Lim X → 0 Tan 3 X − 2 X 3 X − Sin 2 X - Mathematics

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प्रश्न

\[\lim_{x \to 0} \frac{\tan 3x - 2x}{3x - \sin^2 x}\] 

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उत्तर

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

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

 

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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 25 | पृष्ठ ५०

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