Suppose the limit L = lim_{n rightarrow infty | vedclass

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Question

(A) We evaluate the limit $L = \lim_{n \rightarrow \infty} \sqrt{n} \int_0^1 \frac{1}{(1+x^2)^n} dx$. Using the substitution $x = \frac{t}{\sqrt{n}}$,

Vedclass · Accepted Answer

$\frac{1}{2} < L < 2$

Vedclass · Answer

(A) We evaluate the limit $L = \lim_{n \rightarrow \infty} \sqrt{n} \int_0^1 \frac{1}{(1+x^2)^n} dx$. Using the substitution $x = \frac{t}{\sqrt{n}}$,we have $dx = \frac{dt}{\sqrt{n}}$. As $n \rightarrow \infty$,the integral becomes $L = \lim_{n \rightarrow \infty} \sqrt{n} \int_0^{\sqrt{n}} \frac{1}{(1 + t^2/n)^n} \cdot \frac{dt}{\sqrt{n}} = \lim_{n \rightarrow \infty} \int_0^{\sqrt{n}} \left(1 + \frac{t^2}{n}\right)^{-n} dt$. By the definition of the exponential function,$\lim_{n \rightarrow \infty} (1 + t^2/n)^{-n} = e^{-t^2}$. Thus,$L = \int_0^{\infty} e^{-t^2} dt$. We know that the Gaussian integral $\int_0^{\infty} e^{-t^2} dt = \frac{\sqrt{\pi}}{2}$. Since $\pi \approx 3.14159$,$\sqrt{\pi} \approx 1.772$. Therefore,$L = \frac{1.772}{2} = 0.886$. Since $0.5 < 0.886 < 2$,the correct option is $\frac{1}{2} < L < 2$.

Suppose the limit $L = \lim_{n \rightarrow \infty} \sqrt{n} \int_0^1 \frac{1}{(1+x^2)^n} dx$ exists and is larger than $\frac{1}{2}$. Then,

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