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?

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
$$\lim_{n \to \infty} \left(\int_1^2 \frac{x^n}{1+x^n}dx\right) =1$$
