ₙ→∞}≤ft[≤ft(1+{1}{n²})≤ft(1+{4}{n²})≤ft(1+{9}{n²})(2)]¹/n}= [TG EAPCET 9 May 2024 Shift 1]

Correct Answer is : 2e(π−42)2e^{≤ft(π-4/2)}2e(2π−4)

Correct Answer is : 2e(π−42)2e^{≤ft(π-4/2)}2e(2π−4)

Test Index

TS EAMCET 9 May 2024 Shift 1 Solved Paper

Section: Mathematics

Question : 76 of 160

Marks: +1, -0

\lim\limits_{n\to\infty}\left[\left(1+\frac{1}{n^{2}}\right)\left(1+\frac{4}{n^{2}}\right)\left(1+\frac{9}{n^{2}}\right)\ldots(2)\right]^{1/n}=

Solution:

L=\lim\limits_{n\to\infty}\left[\begin{array}{cc}\left(1+\frac{1}{n^{2}}\right) & \left(1+\frac{4}{n^{2}}\right) \\ \left(1+\frac{9}{n^{2}}\right)\ldots(2)\end{array}\right]^{1/n}

=\lim\limits_{n\to\infty}\left(\prod_{K=1}^{n}\left(1+\frac{K^{2}}{n^{2}}\right)\right)^{1/n}

L=\lim\limits_{n\to\infty}e^{\left(\frac{1}{n}\sum\limits_{K=1}^{n}\ln\left(1+\frac{K^{2}}{n^{2}}\right)\right)}\ldots

Sum can be approximated by an integral in the limit as

n\to\infty

\Rightarrow \frac{1}{n}\sum\limits_{K=1}^{n}\ln\left(1+\frac{K^{2}}{n^{2}}\right)\approx \int\limits_{0}^{1}\ln\left(1+x^{2}\right)\,dx

Using integration by parts

\int\limits_{0}^{1}\ln\left(1+x^{2}\right)\,dx=\left[\ln\left(1+x^{2}\right)\cdot x\right]_{0}^{1}

-\int\frac{2x^{2}+2-2}{1+x^{2}}\,dx

=\ln 2-\int\limits_{0}^{1}2\,dx+2\int\limits_{0}^{1}\frac{1}{1+x^{2}}\,dx=\ln 2-2+\frac{\pi}{2}

So, from Eq. (i), we get

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

Go to Question:

Prev Question

Next Question