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

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(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(

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Exists and is $0$

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(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:

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