JEE MATHEMATICS MCQ: LIMITS

1 Easy

1 of 1

Application +4 | -1

\lim\limits_{x \to 0} \frac{{\left( {1 - \cos 2x} \right)\sin 5x}}{{{x^2}\sin 3x}}

equals

View answer and explanation

Correct option: A

\frac{{10}}{3}

Option A (

\frac{{10}}{3}

) is correct because by applying trigonometric identities and standard limits, the expression simplifies to this value.

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\lim \limits_{{x \rightarrow 0}}\left(\frac{1}{x^2}-\operatorname{cosec}^2 x\right)

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\lim \limits_{{x \rightarrow \infty}} x\left(e^{1 / x}-1\right)

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