Evaluate $\lim _{n\to \infty }\int_1^2\:\frac{x^n}{x^n+1}dx$ We have $$I_n=\int _1^2\:\frac{x^n}{x^n+1}dx$$ and we need to find $\lim _{n\to \infty }I_n$. Have any ideea how we can evaluate this limit?
 A: Well, $\frac{x^n}{x^n+1} \leq 1$ for all $x \in [1,2]$, and the constant function $1$ is integrable in $[1,2]$, so by the Dominated Convergence Theorem we have: $$\lim_{n \to +\infty} \int_1^2\frac{x^n}{x^n+1}\,{\rm d}x = \int_1^2 \lim_{n \to +\infty} \frac{x^n}{x^n+1}\,{\rm d}x = \int_{1}^2 1\,{\rm d}x = 1.$$
A: Note that $\frac{x^n}{x^n+1}=1-\frac{1}{x^n+1}$.  Thus,
$$\int_1^2 \frac{x^n}{x^n+1}dx=1-\int_1^2 \frac{1}{1+x^n}dx$$ 
For $n>1$, the integral on the right-hand side satisfies the inequality
$$\left|\int_1^2 \frac{1}{1+x^n}dx\right|\le \int_1^2 x^{-n}dx=\frac{1-2^{1-n}}{n-1}$$
which clearly goes to zero as $n \to \infty$.
Putting all of this together we see that
$$1\ge \int_1^2 \frac{x^n}{1+x^n}dx=1-\int_1^2 \frac{1}{1+x^n}dx \ge 1-\int_1^2 x^{-n}dx=1-\frac{1-2^{1-n}}{n-1}$$
Therefore, by the "squeeze" theorem
$$\lim_{n \to \infty} \left(\int_1^2 \frac{x^n}{1+x^n}dx\right) =1$$
A: Divide the numerator and denominator of the integrand by $x^n$:
$$ \frac{1}{1+x^{-n}} $$
For every $x \in (1,2]$, $x^{-n} \to 0$ as $n \to \infty$. Since this is all but one point of the interval of integration, the integral tends to
$$ \int_1^2 \frac{1}{1} \, dx = 1. $$
To do it more thoroughly than this, you can chop up the interval into a small region around $1$ and the rest, and examine the size of the integrand on each.
