Why $\lim_{n\to\infty}\int_{0}^1f_n(x)dx=\int_{0}^1(\lim_{n\to\infty}f_n(x))dx$? Consider the following equality:
$$\lim_{n\to\infty}\int_{0}^1f_n(x)dx=\int_{0}^1(\lim_{n\to\infty}f_n(x))dx$$
where $$f_n(x):=\frac{x^n}{1+x^n}\qquad x\in [0,1]$$
Since the sequence $(f_n(x))_{n=1}^{\infty}$ is not uniformly convergent, one cannot use the theorem about integration of uniformly convergent function sequences. So here is my question:

How to show that the equality is true?

I think it is equivalent to show that $$\lim_{n\to\infty}\int_{0}^1\frac{1}{1+x^n}dx=1$$
Then things boil down to calculating $$\int_{0}^1\frac{1}{1+x^n}dx$$ for every $n$, which is what I have no idea to do.
 A: Any time you have a sequence of functions bounded in absolute value by the same constant on a finite interval, you can switch the limit and the integral.  (Here they are all bounded above by $1$)
This is a special case of the Dominated Convergence Theorem.
You could also approach your problem more directly.  The limit function is $0$ on $(0,1)$ so that the integral on the right hand side is zero.  Now, we just have to show that $$\lim_{n\rightarrow \infty} \int_0^1 \frac{x^n}{1+x^n}dx=0.$$  Let $\epsilon>0$ be given.  Then consider the decomposition of $[0,1]$ into the intervals $[0,1-\frac{\epsilon}{2}]$ and $[1-\epsilon/2,1]$.  On the second interval, we get an error of at most $\epsilon/2$ since all of our functions are bounded by $1$.  Since the sequence of functions is uniformly convergent on the first interval, we can make it so the integral over that interval is less then $\frac{\epsilon}{2}$ for sufficiently large $n$.  Then all together, this will show that the integral is less then $\epsilon$ for sufficiently large $n$.  This is equivalent to the limit equaling zero.
Hope that helps,
A: Hint:  On the right side, you have a pointwise limit.  It looks like you can show the integral of this limit is zero.  Then I would break the left hand integral at a point depending on $n$, bounding the integral on each side by a decreasing function of $n$ which goes to 0.
A: Lebesgue Dominated Convergence Theorem
A: $$
\int_0^1 {\frac{{x^n }}{{1 + x^n }}dx}  \le \int_0^1 {x^n dx}  = \frac{1}{{n + 1}},
$$
hence the left-hand side integral converges to $0$ as $n \to \infty$. On the other hand, $\frac{{x^n }}{{1 + x^n }} \to 0$ pointwise for $x \in [0,1)$, hence obviously 
$$
\int_0^1 {\bigg(\lim _{n \to \infty } \frac{{x^n }}{{1 + x^n }}\bigg)dx}  = 0.
$$
The key here is that $\frac{{x^n }}{{1 + x^n }} \leq x^n$, $x \in [0,1]$.
