MCQ
Let, $\quad f(x)=\left\{\begin{array}{cc}\frac{x}{\sin x}, & x \in(0,1) \\ 1, & x=0\end{array}\right.$ Consider the integral $I_n=\sqrt{n} \int_0^{1 / n} f(x) e^{-n x} d x$ . Then, $\lim _{n \rightarrow \infty} I_n$
  • A
    Does not exist
  • Exists and is $0$
  • C
    Exists and is $1$
  • D
    Exists and is $1-e^{-1}$

Answer

Correct option: B.
Exists and is $0$
b
(b)

Let $n=\frac{1}{m}$

So, $\lim _{m \rightarrow 0} I_n=\lim _{m \rightarrow 0} \frac{\int_0^m x \cdot e^{-x / m}}{\sin x} d x$

Let $x=m t \Rightarrow d x=m d t$

$\lim _{m \rightarrow 0} I_n=\lim _{m \rightarrow 0} \frac{\int_0^1 \frac{m^2 \cdot e^{-t} \cdot t}{\sin (m t)} d t}{\sqrt{m}}$

$=\lim _{m \rightarrow 0} \sqrt{m} \int_0^1\left(\frac{m t}{\sin (m t)}\right) \cdot e^{-t} d t$

$\lim _{m \rightarrow 0} \sqrt{m} \int_0^1 e^{-t} d t=\lim _{m \rightarrow 0} \sqrt{m}\left(1-e^{-1}\right)=0$

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