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Let $f_n=\frac{\sin(nx)}{n}e^{-x}$. My first guess was to try using Monotone Convergence Theorem but we see that $f_n$ is not monotone increasing. Note that $\lim_{n\rightarrow\infty}f_n=0$ since $\lim_{n\rightarrow\infty}\frac{\sin(n)}n=0$.

I would like to try using Dominated Convergence Thm: Let $g\in L^1(\mathbb{R^n,R})$, $g(x)=e^{-x}$. Then $|f_n|≤g$ a.e. We know $g$ is integrable, in fact $\int^\infty_0e^{-x}=1$. Then by DCT $\lim_{n\rightarrow\infty}\int f_n=\int0=0$

Is it correct? I'm not 100% convinced..

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    $\begingroup$ It is almost correct. Take $f_n$ to be $\frac {\sin (nx)} n 1_{(0,n)}(x) e^{-x}$ $\endgroup$ – Kavi Rama Murthy Oct 21 '20 at 9:33
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Without a convergence theorem:

$| \int_0^n f_n(x) dx| \le \frac{1}{n}\int_0^n e^{-x}dx=\frac{1}{n}(1-\frac{1}{e^n}) \to 0.$

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For more than the limit itself.

Consider $$\int e^{i n x} e^{-x}\,dx=\int e^{i (n+i) x}\,dx=-\frac{i e^{i (n+i) x}}{n+i}$$Now, playing with the complex numbers $$\int \sin(nx)e^{-x}\,dx=-\frac{e^{-x} (\sin (n x)+n \cos (n x))}{n^2+1}$$ $$I_n=\int^n_0\frac1n \sin(nx)e^{-x}\,dx=\frac{1}{n^2+1}-\frac{e^{-n} \left(\sin \left(n^2\right)+n \cos \left(n^2\right)\right)}{n(n^2+1)}$$

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