JEE MATHEMATICS MCQ: LIMITS

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Analyzing +4 | -1

The value of

\lim \limits_{{x \rightarrow \infty}} \frac{2(x)^{1 / 2}+3(x)^{1 / 3}+4(x)^{1 / 4}+\ldots+n(x)^{1 / n}}{(2 x-3)^{1 / 2}+(2 x-3)^{1 / 3}+\ldots+(2 x-3)^{1 / n}}

View answer and explanation

Correct option: A

\sqrt{2}

Option A (

\sqrt{2}

) is correct. By dividing numerator and denominator by

x^{1/2}

and taking the limit as x approaches infinity, the expression simplifies to

\sqrt{2}

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\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}

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