Let f(x) = begin{cases} frac{x}{sin x}, & x i | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $f(x) = \frac{x}{\sin x}$ for $x \in (0, 1)$ and $f(0) = 1$. We want to find $\lim_{n 	o \infty} I_n$ where $I_n = \sqrt{n} \int_0^{1/n} f(

Vedclass · Accepted Answer

Exists and is $0$

Vedclass · Answer

(B) Given $f(x) = \frac{x}{\sin x}$ for $x \in (0, 1)$ and $f(0) = 1$. We want to find $\lim_{n 	o \infty} I_n$ where $I_n = \sqrt{n} \int_0^{1/n} f(x) e^{-nx} dx$. Let $nx = t$,then $x = t/n$ and $dx = dt/n$. As $x$ goes from $0$ to $1/n$,$t$ goes from $0$ to $1$. Substituting these into the integral: $I_n = \sqrt{n} \int_0^1 f(t/n) e^{-t} \frac{dt}{n} = \frac{1}{\sqrt{n}} \int_0^1 \frac{t/n}{\sin(t/n)} e^{-t} dt$. We know that $\lim_{u 	o 0} \frac{u}{\sin u} = 1$. As $n 	o \infty$,$t/n 	o 0$ for $t \in [0, 1]$. Thus,$\lim_{n 	o \infty} \frac{t/n}{\sin(t/n)} = 1$. Using the property of limits under the integral sign: $\lim_{n 	o \infty} I_n = \lim_{n 	o \infty} \frac{1}{\sqrt{n}} \int_0^1 (1) e^{-t} dt = \lim_{n 	o \infty} \frac{1}{\sqrt{n}} [1 - e^{-1}] = 0 	imes (1 - e^{-1}) = 0$.

Let $f(x) = \begin{cases} \frac{x}{\sin x}, & x \in (0, 1) \\ 1, & x = 0 \end{cases}$. Consider the integral $I_n = \sqrt{n} \int_0^{1/n} f(x) e^{-nx} dx$. Then,$\lim_{n \to \infty} I_n$ is:

Similar Questions

$\int_0^a {\frac{{{x^4}\,dx}}{{{{({a^2} + {x^2})}^4}}}} = $

$\int_0^1 {{\cos }^{ - 1}}x\,dx = $

The least value of the function $F(x) = \int_{5\pi /4}^x {(3\sin u + 4\cos u)\,du} $ on the interval $\left[ \frac{5\pi}{4}, \frac{4\pi}{3} \right]$ is

$\int_{-\pi}^{\pi} (\cos px - \sin qx)^2 dx$ is equal to (where $p$ and $q$ are integers).

The value of $\int_{-1 / \sqrt{2}}^{1 / \sqrt{2}}\left(\left(\frac{x+1}{x-1}\right)^{2}+\left(\frac{x-1}{x+1}\right)^{2}-2\right)^{1 / 2} d x$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module