I'm trying to find the following limits using Dominated Convergence Theorem, but can't seem to find a dominating function. Any guidance would be greatly appreciated!
$\lim\limits _{n\rightarrow\infty}\int_{0}^{1}\frac{1+n^2x^2}{(1+x)^{n}}dx$
$\lim_{n\rightarrow\infty}\int_{0}^{\infty}ne^{-nx}\sin(1/x)dx$
For the second problem, it is straightforward to find a (non-negative) dominating function on $[1,\infty)$.
We have $\sin(1/x) > 0$ for $x \in [1,\infty)$ and
$$e^{nx} \geqslant 1 + nx \implies |ne^{-nx}\sin(1/x)| \leqslant\frac{n}{1+nx}\sin(1/x)\leqslant \frac1{x}\sin(1/x),$$
Making the change of variables $u = 1/x$, we see that this choice of dominating function is integrable over $[1,\infty)$:
$$\int_1^{\infty}\frac1{x} \sin(1/x) \, dx= \int_1^{\infty}\frac{ \sin u}{u} \, du < \infty.$$
Since $\displaystyle \lim_{n \to \infty} ne^{-nx}\sin(1/x) = 0$ for all $x > 0$, by the DCT it follows that
$$\lim_{n \to \infty}\int_1^{\infty}ne^{-nx}\sin(1/x) \, dx=0.$$
Alternatively, this holds true for integrals over any interval $[a,\infty)$ with $a > 0$ -- following from the uniform convergence of the sequence of integrands. It is also true with $a= 0$, but it is not likely that this can be demonstrated using the DCT.
For example, there is no dominating integrable function for $ne^{-nx}$, since $\displaystyle \lim_{n \to \infty} ne^{-nx}= 0$ and
$$\lim_{n \to \infty}\int_0^{\infty}ne^{-nx}\, dx = 1.$$
The limit of the integral over $(0,\infty)$ is shown to be $0$ using another approach in: https://math.stackexchange.com/a/59567/148510