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


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?


  • 4
    $\begingroup$ The limit is $1/3$, two different approaches to solve: 1) integrate by parts using $nx^{n-1}=(x^n)'$; 2) substitute $x^n=y$. $\endgroup$
    – metamorphy
    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

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


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