Limit of a sequence of integrals and the integral of a limit For every $n \in \mathbb{Z}_{>0}$ define
$$
f_n(x) = \frac{x^n}{1+x^n}, \, \, \, \, \, x \in [0,1].
$$ I want to show that $$\lim_{n \to \infty} \int_{0}^{1} f_n(x) dx = \int_{0}^{1} \left(\lim_{n \to \infty} f_n(x) \right) dx.$$ I know that the limit function of $f_n$ is
$$
f(x)=\begin{cases}
    0 &: \text{if $x \in [0,1)$}\\
    \frac{1}{2} &: x = 1
  \end{cases}
$$ so I expect the integral on the right to be $0$, but how can I bound the integral on the left appropriately to show this?
EDIT: Also, is there a general principle/theorem I can use to approach problems such as these, for sequences that are not uniformly convergent. For example, are there common ways to bound rational functions of this (or similar) form?

Note that $$\frac{x^n}{1+x^n}\leqslant 2$$ for all $x\in[0,1]$, and $$\int_0^1 2\ \mathsf dx = 2<\infty. $$
So by the dominated convergence theorem,
$$\lim_{n\to\infty}\int_0^1 f_n(x)\ \mathsf dx = \int_0^1 f(x)\ \mathsf dx = 0. $$
