1. ## Help finding limit

Hi

Can anyone please show me how to establish: $\lim_{n \to \infty} \int_0^1\frac{nx^{n-1}}{x+1}dx =1/2$

Thanks a lot.

2. Attempt:

Sub $x = 1 - \frac{t}{n}$

$n dx = - dt$

the integral becomes :

$\lim_{n\to\infty} \int_n^0 \frac{ \left( 1 - \frac{t}{n} \right )^{n-1}}{2 - \frac{t}{n} } ~ ( -dt)$

$= \lim_{n\to\infty} \int_0^n \frac{ \left( 1 - \frac{t}{n} \right )^{n-1}}{2 - \frac{t}{n} } ~ dt$

$= \lim_{n\to\infty} \int_0^n e^{-t} \left [ \frac{1}{2 - \frac{t}{n} } - \frac{1}{2} + \frac{1}{2} \right ] ~dt$

$= \lim_{n\to\infty} \int_0^n e^{-t} \left [\frac{1}{2 - \frac{t}{n} } - \frac{1}{2} \right ]~dt ~+~ \frac{1}{2} \lim_{n\to\infty} \int_0^n e^{-t} ~dt$

$= \lim_{n\to\infty} \int_0^n e^{-t} \cdot {-\frac{t}{n}} \cdot \frac{d}{dy} \left( \frac{1}{y} \right)|_{y\to2} ~ dt ~+~ \frac{1}{2}$

$= \lim_{n\to\infty} \int_0^n e^{-t} \cdot {\frac{t}{n}} \cdot \frac{1}{4}~dt ~+~ \frac{1}{2}$

$= 0 + \frac{1}{2} = \frac{1}{2}$

Another simplier method that i've just thought of :

Suppose the limit exists ,

$I_n = \int_0^1 \frac{ n x^{n-1} }{ 1 + x }~dx$

$= n \int_0^1 \frac{ x^{n-1} + x^{n-2 } - x^{n-2} }{ 1 + x }~dx$

$= n \int_0^1 x^{n-2}~dx - \frac{n}{n-1} I_{n-1}$

$I_n = n \frac{1}{n-1} - \frac{n}{n-1} I_{n-1}$

Let $L = \lim_{n\to\infty} I_n$

we have $L = \lim_{n\to\infty} \frac{n}{n-1} - \lim_{n\to\infty} \frac{n}{n-1} L$

$2L = 1 \implies L = \frac{1}{2}$

3. Fantastic. Thanks a lot for helping me out. Though I understand the simpler solution, I am having trouble parsing the first one. I hope its not too much trouble for you to clarify:

How do you go from

$= \lim_{n\to\infty} \int_0^n \frac{ \left( 1 - \frac{t}{n} \right )^{n-1}}{2 - \frac{t}{n} } ~ dt$

to

$= \lim_{n\to\infty} \int_0^n e^{-t} \left [ \frac{1}{2 - \frac{t}{n} } - \frac{1}{2} + \frac{1}{2} \right ] ~dt$ ?

since $\left( 1 - \frac{t}{n} \right)^{n-1}$ tends to and not equals $e^{-t}$

Thanks again.

4. Perhaps I was wrong , the limit sign is outside the integral

so we cannot put $\left( 1 - \frac{t}{n} \right) ^{n-1} = e^{-t}$

I am here try to make some corrections ,

Put $\frac{1}{ 2 - \frac{t}{n}} = \frac{1}{ 2 - \frac{t}{n}} - \frac{1}{2} + \frac{1}{2}$

$= \left( \frac{1}{ 2 - \frac{t}{n}} - \frac{1}{2} \right ) + \frac{1}{2}$

By MVT, let $f(x) = \frac{1}{x}$

$\frac{1}{ 2 - \frac{t}{n}} - \frac{1}{2} = - [ \frac{1}{2} - \frac{ 1 }{ 2 - \frac{t}{n} } ]$ $= -\frac{t}{n} f'(c) = \frac{t}{n} \cdot \frac{1}{c^2} , 2 - \frac{t}{n} \leq c \leq 2$

we have $\frac{1}{c^2} \leq 1$

$\frac{1}{ 2 - \frac{t}{n}} - \frac{1}{2} \leq \frac{t}{n}$

Therefore ,

$= \lim_{n\to\infty} \int_0^n \frac{ \left( 1 - \frac{t}{n} \right )^{n-1}}{2 - \frac{t}{n} } ~ dt$

$\leq \lim_{n\to\infty} ( \frac{1}{2} ) \int_0^n \left( 1 - \frac{t}{n} \right )^{n-1}~dt ~$ $+~ \lim_{n\to\infty} \int_0^n \frac{t}{n} \left( 1 - \frac{t}{n} \right )^{n-1} ~dt$

In fact ,we don't need to put $\left( 1 - \frac{t}{n} \right )^{n-1} = e^{-t}$

However . after the corrections , I seem to have chosen a terrible detour . Why I thought in this way ...

BTW , just before the bed , I suddenly think of another method , it is also a simple method :

$\lim_{n\to\infty} \int_0^1 \frac{n x^{n-1}}{1 + x}~dx$

Integration by parts ,

$= \frac{x^n}{1 + x} |_0^1 + \int_0^1 \frac{x^n}{(1 + x)^2}~dx$

we just need to prove that the remaining integral tends to zero .

5. $0\le \frac{x^{n}}{(x+1)^{2}}\le x^{n}\implies 0\le \int_{0}^{1}{\frac{x^{n}}{(x+1)^2}\,dx}\le \frac{1}{n+1}.$

the rest follows by the squeeze theorem.

6. Thanks a lot simplependulum and Krizalid for your help!

Also, I realised that the problem can be generalized.

Notice that $\lim_{n \to \infty} n\int_0^1 \left(\frac{1}{1+x} \right)x^ndx = \frac{1}{1+1}$
Also, we observe that \forall polynomial P(x): $\lim_{n \to \infty} n\int_0^1 P(x)x^ndx = P(1)$

Now, let f be continuous on [0,1]. Since, the conditions for Weierstrauss Approximation theorem
are satisfied, we can now establish that $\lim_{n \to \infty} n\int_0^1 f(x)x^ndx = f(1)$

Proof :
Given $f$ and for $\epsilon < 0$ we have some polynomial $P_{\epsilon}$ such that
$P_{\epsilon}(x)-\epsilon < f(x) < P_{\epsilon}(x)+\epsilon \quad \forall x \in [0,1]$
So, $\int_0^1 nP_{\epsilon}(x)x^{n}dx -\epsilon\frac{n}{n+1} < \int_0^1 nf(x)x^{n}dx < \int_0^1 nP_{\epsilon}(x)x^{n}dx +\epsilon\frac{n}{n+1}$
Taking limits we see $P_{\epsilon}(1)-\epsilon < \lim_{n \to \infty} n\int_0^1 f(x)x^ndx < P_{\epsilon}(1)+\epsilon$
That is $lim_{n \to \infty}\int_0^1 nf(x)x^{n}dx$ is $\epsilon$ near $P_{\epsilon}(1)$ and consequently $2\epsilon$ near $f(1)$.
Since $\epsilon$ was arbitrary, we are done.