0
$\begingroup$

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.

$\endgroup$
1
  • 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
2
$\begingroup$

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 $\frac{1}{3}$.

$\endgroup$
2
$\begingroup$

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 $n>1$ $$J_n=\int_0^1\frac{nx^{n-1}}{x+2} dx=\frac 12-\frac n{4(n+1)}\, _2F_1\left(1,n+1;n+2;-\frac{1}{2}\right)$$ $$\lim_{n\to \infty } \, \, _2F_1\left(1,n+1;n+2;-\frac{1}{2}\right)=\frac 23$$ $$J_n \sim \frac{1}{2}-\frac{n}{6 (n+1)}\quad \to\quad \frac 13$$

$\endgroup$
2
$\begingroup$

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

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.