JEE MATHEMATICS MCQ: LIMITS

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Analyzing +4 | -1

The value of

\lim \limits_{{n \rightarrow \infty}} n^{-n^2}\left[(n+1)\left(n+\frac{1}{2}\right)\left(n+\frac{1}{2^2}\right)\right.

\left.\ldots\left(n+\frac{1}{2^{n-1}}\right)\right]^n

View answer and explanation

Correct option: B

e^2

Option B (

e^2

) is correct because the given limit can be simplified to an infinite product of exponentials, which converges to

e^2

after summing the exponents.

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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)

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