Compute the following limit $$ \lim_{n\rightarrow+\infty}n\int_{0}^{1}\frac{x^{n}}{x^{n}+x+1}% \operatorname*{dx}. $$
It is about a $\infty\cdot0$ indeterminate. Denote the expression below the limit with $a_{n}$. I just noted that if $$ 0\leq x\leq1\Rightarrow0\leq x^{n}\leq1 \quad \text{and}% \quad 1\leq x^{n}+x+1\leq3; $$ hence $$ \frac{1}{3}\leq\frac{1}{x^{n}+x+1}\leq1\quad \text{or} \quad \frac{x^{n}}{3}\leq\frac{x^{n}}{x^{n}+x+1}\leq x^{n}; $$ then I integrated the above inequality to get that% $$ \forall n\in \mathbb{N} :\frac{1}{3n+3}\leq\frac{a_{n}}{n}\leq\frac{1}{n+1}. $$ So $\lim\limits_{n\rightarrow+\infty}\frac{a_{n}}{n}=0$. It is possible now to compute the limit of $\left( a_{n}\right) $ from the previous limit? Any other way to compute the limit? Thank you.
The change of variables $y=x^n$ removes the external factor of $n$, putting all the dependence inside the integral: $$ n\int_{0}^{1}\frac{x^{n}}{x^{n}+x+1} \,dx = n \int_{0}^{1}\frac{y}{y+y^{1/n}+1} \tfrac1ny^{1/n-1} dy = \int_{0}^{1}\frac{y}{y^{2-1/n}+y+y^{1-1/n}} \,dy. $$ The integrand is an increasing function of $n$, so by the monotone convergence theorem, $$ \lim_{n\to\infty} \int_{0}^{1}\frac{y}{y^{2-1/n}+y+y^{1-1/n}} \,dy = \int_{0}^{1}\frac{y}{y^2+y+y^1} \,dy = \int_{0}^{1}\frac{1}{y+2} \,dy = \log\frac32. $$