JEE MATHEMATICS MCQ: LIMITS

1 Easy

1 of 1

Analyzing +4 | -1

f\left( x \right)

is the integral function of the function

\frac{{2\sin x - \sin 2x}}{{{x^3}}},x \ne 0

, then

{\lim _ \to }f'\left( x \right)

is equal to

View answer and explanation

Correct option: B

Option B (1) is correct because it is the result of evaluating the limit of the derivative of the integral function.

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

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f^{\prime}(0)

\lim\limits_{n \to \infty } {\left( {1 + \sin \frac{a}{n}} \right)^n}