ₙ→∞} [≤ft(1+1/n²) × ≤ft(1+2²/n²) × ≤ft(1+n²/n²)]¹/n} =

Correct Answer is : 2eπ−422 e^{π-4/2}2e2π−4

Correct Answer is : 2eπ−422 e^{π-4/2}2e2π−4

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AP EAMCET Engineering 2018 Apr 24 Shift 1 Paper

Section: Mathematics

Question : 76 of 160

Marks: +1, -0

\lim\limits_{n\to\infty}

[\left(1+\frac{1}{n^2}\right)

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

\left(1+\frac{n^2}{n^2}\right)]^{1/n}

Solution:

Consider

A =

\lim\limits_{n\to\infty}

[\left(1+\frac{1}{n^2}\right)

\left(1+\frac{2^2}{n^2}\right) \ldots \times \left(1+\frac{n^2}{n^2}\right)]^{1/n}

Take log on both sides

\log A = \lim\limits_{n \to \infty} \frac{1}{n} [\log \left(1+\frac{1}{n^2}\right) + \log \left(1+\frac{2^2}{n^2}\right)

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

\log A = \lim\limits_{n \to \infty} \sum\limits_{r=1}^{n} \log \left(1+\frac{r^2}{n^2}\right) \frac{1}{n}

= \int\limits_{0}^{1} \log (1+x^2) dx

Using integration by parts

\log A = \log (1+x^2) \int 1 dx

- \int \frac{d}{dx} (\log (1+x^2)) \int 1 dx

= x \log (1+x^2) - \int \frac{2x}{1+x^2} x dx

= x \log (1+x^2) - 2 \int \frac{x^2}{1+x^2} dx

= \left[ x \log (1+x^2) - 2x + 2 \tan^{-1}(x) \right]_{0}^{1}

\log A = \left[ \log 2 - 2 + 2 \tan^{-1}(1) \right]

A = e^{\log 2 - 2 + 2 \left( \frac{\pi}{4} \right)} = 2 e^{\frac{\pi-4}{2}}

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