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Lim X → 0 ( 1 − Cos 2 X ) Sin 5 X X 2 Sin 3 X =

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प्रश्न

\[\lim_{x \to 0}  \frac{\left( 1 - \cos 2x \right) \sin 5x}{x^2 \sin 3x} =\]

विकल्प

  • (a) 10/3 

  • (b) 3/10 

  • (c) 6/5 

  • (d) 5/6

MCQ
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उत्तर

(a)  10/3 

\[\lim_{x \to 0} \frac{\left( 1 - \cos 2x \right) \sin 5x}{x^2 \sin 3x}\]
\[ = \lim_{x \to 0} \frac{2 \sin^2 x \sin 5x}{x^2 \sin 3x}\]
\[ = \lim_{x \to 0} \frac{\frac{2 \sin^2 x}{x^2} \left( \frac{\sin 5x}{5x} \right) \times 5}{\left( \frac{\sin 3x}{3x} \right)3}\]
\[ = \frac{10 \times 1^2 \times 1}{1 \times 3}\]
\[ = \frac{10}{3}\]

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अध्याय 29: Limits - Exercise 29.13 [पृष्ठ ७८]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 29 Limits
Exercise 29.13 | Q 5 | पृष्ठ ७८

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