JEE MATHEMATICS MCQ: LIMITS

1 Easy

1 of 1

Analyzing +4 | -1

The value of

\lim \limits_{{n \rightarrow \infty}} \sum\limits_{{k=1}}^n \frac{n-K}{n^2} \cos \left(\frac{4 K}{n}\right)

equals

View answer and explanation

Correct option: D

\frac{1}{16}(1-\cos 4)

The correct answer is D (

\frac{{1}}{{16}}(1-\cos 4)

) because the limit of the Riemann sum can be expressed as a definite integral, which can then be evaluated using integration by parts.

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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_{h \to 0} 2\left\{ {\frac{{\sqrt 3 \sin \left( {\frac{\pi }{6} + h} \right) - \cos \left( {\frac{\pi }{6} + h} \right)}}{{\sqrt 3 h\left( {\sqrt 3 \cos \,h - \sin \,h} \right)}}} \right\}

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

f(x)=\frac{(x+1)^7 \sqrt{1+x^2}}{\left(x^2-x+1\right)^6}

f^{\prime}(0)

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