# Hard limit proof problem

#### Cephal

##### New member
Hello everybody,

I have proved that:

$$\displaystyle \displaystyle \lim_{n\to+\infty} n\int_{0}^1 x^ng(x)\mathrm{d}x=g(1)$$
with $$\displaystyle g \in\mathcal{C}^0(\left[0,1\right],\mathbb{R})$$.

But I don't know how to prove this:

$$\displaystyle \displaystyle \int_0^1 x^n f(x)\mathrm{d}x=\dfrac{f(1)}{n}-\dfrac{f(1)+f'(1)}{n^2}+o_{+\infty}(\dfrac{1}{n^2})$$
with $$\displaystyle f\in\mathcal{C}^1(\left[0,1\right],\mathbb{R})$$.

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#### Opalg

##### MHB Oldtimer
Staff member
Hello everybody,

I have prove that:

$$\displaystyle \displaystyle \lim_{n\to+\infty} n\int_{0}^1 x^ng(x)\mathrm{d}x=g(1)$$
with $$\displaystyle g \in\mathcal{C}^0(\left[0,1\right],\mathbb{R})$$.

But I don't know how to prove this:

$$\displaystyle \displaystyle \int_0^1 x^n f(x)\mathrm{d}x=\dfrac{f(1)}{n}-\dfrac{f(1)+f'(1)}{n^2}+o_{+\infty}(\dfrac{1}{n^2})$$
with $$\displaystyle f\in\mathcal{C}^1(\left[0,1\right],\mathbb{R})$$.

Hi Cephal, and welcome to MHB!

What you know about $g$ is (1) $g$ is continuous and therefore bounded on $[0,1]$, say $|g(x)| \leqslant M$ for some $M$; (2) $g(x)$ is continuous at $x=1$ and therefore given $\varepsilon>0$ there exists $\delta>0$ such that $|g(x) - g(1)| < \varepsilon$ whenever $1- \delta \leqslant x \leqslant 1$.

Now write $$\displaystyle n\int_0^1 x^n g(x)\,dx = n\int_0^{1-\delta} x^n g(x)\,dx + n\int_{1-\delta}^1 x^n g(x)\,dx.$$ Show that if $n$ is large then the first of those integrals is small and the second one is close to $g(1)$.

#### Cephal

##### New member

In fact I have already proved that:

$$\displaystyle \displaystyle \lim_{n\to+\infty} n\int_{0}^1 x^ng(x)\mathrm{d}x=g(1)$$
with $$\displaystyle g \in\mathcal{C}^0(\left[0,1\right],\mathbb{R})$$.

My problem is to prove this:

$$\displaystyle \displaystyle \int_0^1 x^n f(x)\mathrm{d}x=\dfrac{f(1)}{n}-\dfrac{f(1)+f'(1)}{n^2}+o_{+\infty}(\dfrac{1}{n^2})$$
with $$\displaystyle f\in\mathcal{C}^1(\left[0,1\right],\mathbb{R})$$.

#### chisigma

##### Well-known member
Hello everybody,

I have proved that:

$$\displaystyle \displaystyle \lim_{n\to+\infty} n\int_{0}^1 x^ng(x)\mathrm{d}x=g(1)$$
with $$\displaystyle g \in\mathcal{C}^0(\left[0,1\right],\mathbb{R})$$.

But I don't know how to prove this:

$$\displaystyle \displaystyle \int_0^1 x^n f(x)\mathrm{d}x=\dfrac{f(1)}{n}-\dfrac{f(1)+f'(1)}{n^2}+o_{+\infty}(\dfrac{1}{n^2})$$
with $$\displaystyle f\in\mathcal{C}^1(\left[0,1\right],\mathbb{R})$$.

You can proceed with integration by parts obtaining...

$\displaystyle \int_{0}^{1} x^{n}\ f(x)\ dx = \frac{f(1)}{n+1} - \frac{1}{n+1}\ \int_{0}^{1} x^{n + 1}\ f^{\ '} (x)\ dx =$

$\displaystyle = \frac{f(1)}{n+1} - \frac{f^{\ '}(1)}{(n+1) (n+2)} + \frac{1}{(n+1) (n+2)}\ \int_{0}^{1} x^{n+2}\ f^{\ ''} (x)\ d x\ (1)$

Now if You consider that...

$\displaystyle \frac{f(1)}{n + 1} = \frac{f(1)}{n} - \frac{f(1)}{n^{2}} + \mathcal{o}\ (\frac{1}{n^{2}})\ (2)$

... and...

$\displaystyle \frac{f^{\ '}(1)}{(n+1) (n+2)} = \frac{f^{\ '}(1)}{n^{2}} + \mathcal{o}\ (\frac{1}{n^{2}})\ (3)$

... You arrive to the result...

Kind regards

$\chi$ $\sigma$

#### Cephal

##### New member
You can proceed with integration by parts obtaining...

$\displaystyle \int_{0}^{1} x^{n}\ f(x)\ dx = \frac{f(1)}{n+1} - \frac{1}{n+1}\ \int_{0}^{1} x^{n + 1}\ f^{\ '} (x)\ dx =$

$\displaystyle = \frac{f(1)}{n+1} - \frac{f^{\ '}(1)}{(n+1) (n+2)} + \frac{1}{(n+1) (n+2)}\ \int_{0}^{1} x^{n+2}\ f^{\ ''} (x)\ d x\ (1)$

Now if You consider that...

$\displaystyle \frac{f(1)}{n + 1} = \frac{f(1)}{n} - \frac{f(1)}{n^{2}} + \mathcal{o}\ (\frac{1}{n^{2}})\ (2)$

... and...

$\displaystyle \frac{f^{\ '}(1)}{(n+1) (n+2)} = \frac{f^{\ '}(1)}{n^{2}} + \mathcal{o}\ (\frac{1}{n^{2}})\ (3)$

... You arrive to the result...

Kind regards

$\chi$ $\sigma$
Thank you very very much $$\displaystyle \chi\sigma$$.

#### Opalg

##### MHB Oldtimer
Staff member
Apologies for not reading the original question carefully enough. Chisigma's approach is certainly the right way to go. My only comment on it is that you are only told that $$\displaystyle f\in\mathcal{C}^1(\left[0,1\right],\mathbb{R})$$. So you should not assume that $f$ is twice differentiable. Instead, use chisigma's method to integrate by parts just once, getting $$\displaystyle \int_{0}^{1} x^{n}f(x)\, dx = \frac{f(1)}{n+1} - \frac{1}{n+1} \int_{0}^{1} x^{n + 1} f^{\, '} (x)\, dx.$$ Then write that as $$\displaystyle \int_{0}^{1} x^{n}f(x)\, dx = \frac{f(1)}{n+1} - \frac{1}{(n+1)^2}\left[(n+1) \int_{0}^{1} x^{n + 1} f^{\, '} (x)\, dx\right],$$ and use the previous result about $g(x)$ to conclude that the part in the large brackets is $f^{\, '} (1) + o_{n\to\infty}(1).$

#### chisigma

##### Well-known member
Now is my turn to apologies because I have had a little hurry in my last post, so that some essential step has been omitted. The most important step I have omitted is that continuity is not neccesary condition for integrability of a function but is necessary condition for univocal definition of a function in a closed interval. Now if we write again...

$\displaystyle \int_{0}^{1} x^{n}\ f(x)\ d x = \frac{f(1)}{n+1} - \frac{f^{\ '}(1)}{(n + 1) (n + 2)} + \frac{1}{(n+1) (n+2)}\ \int_{0}^{1} x^{n+2} f^{\ ''}(x)\ dx\ (1)$

... taking into account the relations...

$\displaystyle \frac{1}{n + 1} = \frac{1}{n}\ (1 - \frac{1}{n} + \frac{1}{n^{2}} - ...)\ (2)$

$\displaystyle \frac{1}{n+2} = \frac{1}{n}\ (1 - \frac{2}{n} + \frac{4}{n^{2}} - ...)\ (3)$

all that we can say is...

a) if f(*) is continuous in [0,1] then f(*) is 'well defined' in x=1 and is...

$\displaystyle \int_{0}^{1} x^{n}\ f(x)\ dx = \frac{f(1)}{n} + \mathcal {o}\ (\frac{1}{n})\ (4)$

b) if f(*) and its derivative are continous in [0,1] then f(*) and its derivative are 'well defined' in x=1 and is...

$\displaystyle \int_{0}^{1} x^{n}\ f(x)\ dx = \frac{f(1)}{n} - \frac{f(1) + f^{\ '}(1)}{n^{2}} + \mathcal {o}\ (\frac{1}{n^{2}})\ (5)$

Of course if all the derivatives of f(x) in x=1 are 'well defined', then we can extend the expansion without limits...

Kind regards

$\chi$ $\sigma$