To evaluate limit $\lim_{n \to \infty} (n+1)\int_0^1x^{n}f(x)dx$

Question

Let $f:\Bbb{R} \rightarrow \Bbb{R}$ be a differentiable function whose derivative is continuous, then

$$\lim_{n \to \infty} \left((n+1)\int_0^1x^{n}f(x)dx\right).$$

I think I have to use L'Hop rule, but I don't see how.

Answer 1

You may just integrate by parts: $$ (n+1)\int_0^1x^{n}f(x)dx=\left.x^{n+1}f(x)\right|_0^1-\int_0^1x^{n+1}f'(x)dx=f(1)-\int_0^1x^{n+1}f'(x)dx $$ and use the fact that $f'$ is continuous over $[0,1]$ giving $$ |f'(x)|\leq M, \quad x \in [0,1], $$ and $$ \left|\int_0^1x^{n+1}f'(x)dx\right|\leq\int_0^1x^{n+1}|f'(x)|dx\leq\frac{M}{n+2} $$ to conclude.

Answer 2

We can prove that it holds for $f$ is continuous. ($f$ does not need to be differentiable.) $$ \lim_{n \to \infty} (n+1)\int_0^1x^{n}f(x)dx=f(1) $$ Since $f$ is continuous, $f$ is bounded on $[0,1]$. Also $$ \forall \epsilon>0, \exists \delta>0, \forall x\in(1-\delta,1],\text{there is } |f(x)-f(1)|<\epsilon $$

\begin{align} \left|(n+1)\int_{0}^{1}x^nf(x)dx-f(1)\right|&= \left|\int_{0}^{1-\delta} (n+1)x^nf(x)dx+\int_{1-\delta}^{1} (n+1)x^nf(x)dx-f(1)\right| \\ &=\left|f(t_1) \int_{0}^{1-\delta} (n+1)x^ndx+f(t_2)\int_{1-\delta}^{1} (n+1)x^ndx-f(1)\right| \hspace{5 mm} \\ &\hspace{10 mm}(t_1\in(0,1-\delta),t_2\in(1-\delta,1) \text{ and by IMVT}) \\ &=\left|f(t_1)(1-\delta)^{n+1}+f(t_2)(1-(1-\delta)^{n+1})-f(1)\right| \\ &\leqslant 2M(1-\delta)^{n+1}+|f(t_2)-f(1)| \\&<2M(1-\delta)^{n+1}+\epsilon \end{align} So $$ \varlimsup\limits_{n\to\infty}\left|(n+1)\int_{0}^{1}x^nf(x)dx-f(1)\right|\leqslant \varlimsup\limits_{n\to\infty}2M(1-\delta)^{n+1}+\epsilon=\epsilon $$ Since $\epsilon$ is arbitrary small $$ \varlimsup\limits_{n\to\infty}\left|(n+1)\int_{0}^{1}x^nf(x)dx-f(1)\right|=0 \hspace{5 mm} \text{and}\hspace{5 mm} \varliminf\limits_{n\to\infty}\left|(n+1)\int_{0}^{1}x^nf(x)dx-f(1)\right|=0 $$

So $$ \lim\limits_{n\to\infty}\left|(n+1)\int_{0}^{1}x^nf(x)dx-f(1)\right|=0\hspace{5 mm} \text{or} \hspace{5 mm}\lim\limits_{n\to\infty}(n+1)\int_{0}^{1}x^nf(x)dx=f(1) $$

Answer 3

By condition and integration by parts formula, it follows that \begin{align} & (n + 1)\int_0^1 x^n f(x) dx \\ = & \int_0^1 f(x) d(x^{n + 1}) \\ = & \left.f(x)x^{n + 1} \right|_0^1 - \int_0^1 x^{n + 1}f'(x)dx \\ = & f(1) - \int_0^1 x^{n + 1}f'(x) dx. \end{align} Since $f'$ is continuous, $f'$ is bounded on the interval $[0, 1]$, assume it is bounded by $M \geq 0$, then, \begin{align} & \left|\int_0^1 x^{n + 1}f'(x) dx\right| \\ \leq & \int_0^1 x^{n + 1}|f'(x)| dx \\ \leq & M\int_0^1 x^{n + 1} dx \\ = & \frac{M}{n + 2} \to 0 \end{align} as $n \to \infty$.

Hence the result of the original limit is $f(1)$.

Answer 4

If $f(x)=x^m$ then the limit is $1$. This is easy to check. Similarly, if $f$ is a polynomial, one uses above to show that the limit if $f(1)$. Now, polynomials are dense in $C[0,1]$, hence for given continous $f$ and $\epsilon$ one can find a polynomial such that absolutely value of the difference between $f$ and $g$ is less then $\epsilon$. The integral of the difference is not bigger than $1$ easy to check, and as noted above, for polynomial, the lim8t is its value at $1$. Thus, the limit is always equal to the value of function at $1$.