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प्रश्न
The value of \[\lim_{x \to 0} \frac{1 - \cos x + 2 \sin x - \sin^3 x - x^2 + 3 x^4}{\tan^3 x - 6 \sin^2 x + x - 5 x^3}\] is
पर्याय
1
2
−1
−2
उत्तर
(b) 2
\[\lim_{x \to 0} \frac{\left( 1 - \cos x \right) + 2 \sin x - \sin^3 x - x^2 + 3 x^4}{\tan^3 x - 6 \sin^2 x + x - 5 x^3}\]
\[\text{ Dividing } N^r \text{ and } D^r \text{ by } x: \]
\[ \lim_{x \to 0} \frac{\frac{2 \sin^2 \frac{x}{2}}{x} + \frac{2 \sin x}{x} - \frac{\sin^3 x}{x} - x + 3 x^3}{\frac{\tan^3 x}{x} - \frac{6 \sin^2 x}{x} + 1 - 5 x^2}\]
\[ \lim_{x \to 0} \frac{\frac{2x}{4} \frac{\sin^2 \frac{x}{2}}{\frac{x^2}{4}} + \frac{2\sin x}{x} - \frac{x^2 \sin^3 x}{x^3} - x + 3 x^3}{x^2 \left( \frac{\tan 3x}{x^3} \right) - \frac{6x \sin^2 x}{x^2} + 1 - 5 x^2}\]
\[ = \frac{0 + 2 - 0 - 0 + 0}{0 - 0 + 1 - 0} = 2\]
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