Evaluating $ \lim_{n\rightarrow\infty} n \int_{0}^{1} \frac{{x}^{n-2}}{{x}^{2n}+x^n+1} \mbox {d}x$ Evaluating
$$L = \lim_{n\rightarrow\infty} n \int_{0}^{1} \frac{{x}^{n-2}}{{x}^{2n}+x^n+1} \mbox {d}x$$
 A: Marvis showed that
$$
I_n = \int_0^1 \dfrac{nx^{n-2}}{x^{2n} + x^n + 1} dx = \int_0^1 \dfrac{dt}{t^{1/n}(t^2 + t + 1)}.
$$
At this point one can use monotone convergence as Davide commented, but since Chris is interested in a more direct approach, here is one.
We have
$$
I_n \geq \int_0^1 \dfrac{dt}{t^2 + t + 1}=\frac{\pi}{3\sqrt3},
$$
and for $n>1$ and for any small $\delta>0$
$$
I_n\leq \int_0^\delta \dfrac{dt}{t^{1/n}} + \int_\delta^1 \dfrac{dt}{\delta^{1/n}(t^2 + t + 1)}=\frac{\delta^{1-1/n}}{1-1/n}+\frac1{\delta^{1/n}}\Big(\frac{2\pi}{3\sqrt3}-\frac{2\arctan(\frac{2\delta+1}{\sqrt3})}{\sqrt3}\Big).
$$
An inspection of the limit as $n\to\infty$ of the right hand side shows that
$$
I_n\leq \delta+\frac{2\pi}{3\sqrt3}-\frac{2\arctan(\frac{2\delta+1}{\sqrt3})}{\sqrt3}+E(\delta,n),
$$
where $E(\delta,n)\to0$ as $n\to\infty$ for any fixed $\delta>0$. Since $\arctan(z)$ is continuous at $z=\frac1{\sqrt3}$ with $\arctan(\frac1{\sqrt3})=\frac\pi6$, for any given $\varepsilon>0$, we can choose $\delta>0$ so small that
$$
\delta+\frac{2\pi}{3\sqrt3}-\frac{2\arctan(\frac{2\delta+1}{\sqrt3})}{\sqrt3}<\frac{\pi}{3\sqrt3}+\frac\varepsilon2.
$$
For sufficiently large $n$, we can also ensure $E(\delta,n)<\frac\varepsilon2$, implying that there is a threshold value $N$ such that
$$
\frac{\pi}{3\sqrt3}\leq I_n<\frac{\pi}{3\sqrt3}+\varepsilon,
$$
for all $n>N$.
A: Let $x^n = t$. Then $n x^{n-1} dx = dt \implies nx^{n-2} dx = \dfrac{dt}x = \dfrac{dt}{t^{1/n}}$.
$$I_n = \int_0^1 \dfrac{nx^{n-2}}{x^{2n} + x^n + 1} dx = \int_0^1 \dfrac{dt}{t^{1/n}(t^2 + t + 1)}$$
Hence, $$\lim_{n \rightarrow \infty} I_n = \int_0^1 \dfrac{dt}{t^2 + t + 1} = \dfrac{\pi}{3 \sqrt{3}}$$
But can I change the limit and integral?...
