Calculate $\lim_{n\to\infty}\int_0^\frac{1}{n} \frac{e^{-t} \sin{\frac{t}{n}} dt}{(1+\frac{t}{n}^2)\arctan{\frac{t}{n}}}$ How to find 

$$\lim_{n\to\infty}\int_0^\frac{1}{n} \frac{e^{-t} \sin{\frac{t}{n}} dt}{(1+\frac{t}{n}^2)\arctan{\frac{t}{n}}}?$$

Any help is welcome.
Thanks in advance. 
 A: Notice that the function $g(x)=\left|\frac{\sin x}{\arctan x}\right|$ is bounded by some $C$ on the interval $[0,1]$, so 
$$\left|\frac{e^{-t}\sin\frac{t}{n}}{\left(1+\frac{t^2}n\right)\arctan \frac{t}{n}}\right| \le C$$ 
Therefore, 
$$\left|\int_0^{1/n}\frac{e^{-t}\sin\frac{t}{n}}{\left(1+\frac{t^2}n\right)\arctan \frac{t}{n}}\ dt\right| \le \frac{C}{n}.$$
This means that the desired limit is $0$.
Comment:  When you change the upper bound of the integral from $1/n$ to $n$, the problem becomes more interesting. 
A: Let $f_n= \frac{e^{-t} \sin{\frac{t}{n}} dt}{(1+\frac{t}{n}^2)\arctan{\frac{t}{n}}}1_{[0,\frac{1}{n}]}$.
We have that $f_n(t) \to 0$ since $g(tx)=\frac{\sin{tx}}{\arctan{tx}} \to t^2$  as $x \to 0$ thus $0 \leq |g(\frac{t}{n})| \leq  (1+\frac{1}{n^2}) \leq 2$ for all $t \in (0,\delta_n]$ for some $\frac{1}{2}>\delta_n >0$ 
We can choose $\delta_n$ such that $\delta _1>\delta_2>...>\delta_n>...$
Thus $(0,\delta_n] \subseteq (0,\delta_1],\forall n \in \Bbb{N}$ so $|g(\frac{t}{n})| \leq 2 ,\forall n \in \Bbb{N}$
Also $|g(\frac{t}{n})|e^{-t} \leq M e^{-t}, \forall x \in [\delta_0,+\infty]$ for some $M>0$ independent of $n$ 
Thus  $|f_n(t)| \leq 2e^{-t}+Me^{-t} \in L^1[0,+\infty]$ because $1_{[\delta_0,+\infty]},1_{(0,\delta_0]} \leq 1$
So by dominated convergence ,the limit is zero.
