# Evaluate $\lim_{n \rightarrow \infty} \int_0^1 \frac{nx^{n-1}}{2+x} dx$

Evaluate $$\lim_{n \rightarrow \infty} \int_0^1 \frac{nx^{n-1}}{2+x} dx$$.

Attempt:

Let $$h(x)=x^n$$. Then $$nx^{n-1} = h'(x)$$.

I'm thinking about using the Dominated Convergence Theorem.

So $$f_n= \frac{nx^{n-1}}{2+x}$$.

But I'm not sure what the limit of $$f_n$$. And I couldn't find any bounds for $$f_n$$.

So I don't know if DCT will work here.

If DCT doesn't apply here, what would be a good way to solve this?

Thanks.

• The limit is $1/3$, two different approaches to solve: 1) integrate by parts using $nx^{n-1}=(x^n)'$; 2) substitute $x^n=y$. Feb 7 at 23:55

tl;dr: you can use the DCT after the change of variable.

If you want to use the DCT (it's overkill, but why not?): first, do the change of variable $$u = x^{n}$$ you were thinking of. Then, $$\int_0^1 \frac{n x^{n-1}}{2+x}dx = \int_0^1 \frac{1}{2+u^{1/n}}du$$ and you can now apply the DCT to the new integral (the dominating function being, for instance, $$g=1/2$$). This gives you $$\int_0^1 \frac{n x^{n-1}}{2+x}dx = \int_0^1 \frac{1}{2+u^{1/n}}du \xrightarrow[n\to\infty]{} \int_0^1 \frac{1}{3}du = \frac{1}{3}$$ since the pointwise limit of $$f_n(u)=\frac{1}{2+u^{1/n}}$$ on $$(0,1]$$ is $$\frac{1}{3}$$.

The integrand strongly suggest a gaussian hypergeometric function with a last argument being $$\pm \frac x 2$$ $$I_n=\int\frac{nx^{n-1}}{x+2} dx=\frac{x^n}{2}-\frac{n\, x^{n+1} }{4 (n+1)}\, _2F_1\left(1,n+1;n+2;-\frac{x}{2}\right)$$ Assuming $$n>1$$ $$J_n=\int_0^1\frac{nx^{n-1}}{x+2} dx=\frac 12-\frac n{4(n+1)}\, _2F_1\left(1,n+1;n+2;-\frac{1}{2}\right)$$ $$\lim_{n\to \infty } \, \, _2F_1\left(1,n+1;n+2;-\frac{1}{2}\right)=\frac 23$$ $$J_n \sim \frac{1}{2}-\frac{n}{6 (n+1)}\quad \to\quad \frac 13$$

I think at least one of the suggested ways in metamorphy's comment should appear as answer:

Integration by parts gives:

$$\begin{eqnarray*}\int_0^1 \frac{nx^{n-1}}{2+x}dx & = & \left.\frac{x^{n}}{2+x}\right|_0^1 + \underbrace{\int_0^1\frac{x^{n}}{(2+x)^2}dx}_{I_n} \\ & = & \frac 13 + I_n \end{eqnarray*}$$

Now, without any dominated convergence you get

$$0\leq I_n \leq \frac 14\int_0^1x^n\;dx = \frac 1{4(n+1)}\stackrel{n\to\infty}{\longrightarrow}0$$