Evaluate $\lim_{n\to\infty}n^2\int_0^1x^nf(x)dx$ if $f(1)=0$.
I know that if $f$ is continuous then $\int_0^1x^nf(x)dx=0$ by applying substitution $u=x^n$ and using the same substitution $n\times\int_0^1 x^nf(x)dx \to f(1)$ as $n\to\infty$.But what happens if i still have an $n$left in the front from $n^2$ and $f(1)=0$. My particular case is: $f(x)=(x-1)\times e^{-\frac{1}{x+5}}$. Any hints?
If $f \in C^1 ([0,1])$ (as in your particular case), we can integrate by parts to find $$I_n \equiv n^2 \int \limits_0^1 x^n f(x) \, \mathrm{d} x = - \frac{n^2}{n+1} \int \limits_0^1 x^{n+1} f'(x) \, \mathrm{d} x \, , \, n \in \mathbb{N} \, .$$ Then the substitution $ t = x^{n+2}$ yields $$ I_n = -\frac{n^2}{(n+1)(n+2)} \int \limits_0^1 f'\left(t^{\frac{1}{n+2}}\right) \, \mathrm{d} x \, , \, n \in \mathbb{N} \, .$$ Now the prefactor converges to $1$ as $n \to \infty$ . Since $f'$ is continuous, the integrand converges to $f'(1)$ for every $t \in (0,1]$ . Since it is also bounded, we can use the dominated convergence theorem to conclude $$ \lim_{n \to \infty} I_n = - \left[\lim_{n \to \infty} \frac{n^2}{(n+1)(n+2)} \right] \int \limits_0^1 \lim_{n \to \infty} f'\left(t^{\frac{1}{n+2}}\right) \, \mathrm{d} t = - 1 \cdot \int \limits_0^1 f'(1) \, \mathrm{d} t = - f'(1) \, . $$