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Consider the function $f$ which is continuous. Calculate $\lim_{n \to \infty} \int_0^1nx^nf(x)dx$.

Here first I attempted to prove $f_n=nx^n$ is uniformly convergent using sup-norm limit but unfortunately, it is not uniformly convergent, as $M_n=sup_{x\in[0,1]}f_n(x)=1$ so $M_n \not\to 0$. So, I guess that the limit would not be zero.

I have an intuition that it would be $1$, but can't prove it. Any help!!

Ri-Li
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  • It cannot be $1$ for every $f$. If you change $f$ to $2f$ the limit should get multiplied by $2$ right? – Kavi Rama Murthy May 13 '19 at 07:37
  • If $f(x)=x^m$ for some $m\in\mathbb{Z}{\geq0}$, then the integral (depending on $n$) is $\int_0^1nx^{m+n}dx=\left[\frac{n}{m+n+1}x^{m+n+1}\right]_0^1=n/(m+n+1)\to 1$ as $n\to\infty$. Now, for general continuous $f$, we can write out its Taylor series $f(x)=\sum{n\geq0}a_nx^n$ so the limit taking general $f$ is $\sum_{n\geq0}a_01^n=f(1)$. There are rules for infinite sums inside limits and integrals that justify this reasoning. – RMWGNE96 May 13 '19 at 08:32
  • @Jean-ClaudeArbaut . Exact duplicate . I have added a new A there, which is essentially the same proof as the other detailed A but in a different style. The 3-line A there, using the Weierstrass Theorem, is also worth thinking about. – DanielWainfleet May 13 '19 at 11:17

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