Indirect application of dominated convergence theorem Let $f:[0,1]\to \mathbb{C}$ be Lebesgue integrable and be continuous at $1$. Prove that
$$\lim_{n\to \infty} n\int_0^1 f(x)x^n dx=f(1)$$
My attempt:
In order to use the fact that $f$ is continuous at $1$ i.e. $|x-1|<\delta \implies |f(x)-f(1)|<\epsilon$. So if $x\in (1-\delta,1)$, then either $f(x)<f(1)+\epsilon$ or $f(x)>f(1)-\epsilon$. In the following split up of integral, I am considering $f(x)<f(1)+\epsilon$:
\begin{align*}
\lim_{n\to \infty} n\int_0^1 f(x)x^n dx &=\lim_{n\to \infty} n\int_0^{1-\delta} f(x)x^n dx + \lim_{n\to \infty}n\int_{1-\delta}^1 f(x)x^n dx \\
&< \lim_{n\to \infty} n\int_0^{1-\delta} f(x)x^n dx + (f(1)+\epsilon) \lim_{n\to \infty} n\int_{1-\delta}^1 x^n dx \\
&= \lim_{n\to \infty} n\int_0^{1-\delta} f(x)x^n dx + (f(1)+\epsilon) \lim_{n\to \infty} n \left[\frac{x^{n+1}}{n+1}\right]_{1-\delta}^1 \\
&= \lim_{n\to \infty} n\int_0^{1-\delta} f(x)x^n dx + (f(1)+\epsilon)
\end{align*}
But I'm not sure how to proceed and apply LDCT?

On $(0,1-\delta)$ we have $|nf(x)x^{n}|\leq n(1-\delta)^{n} |f(x)|$. Note that $n(1-\delta)^{n} \to 0$ as $ n \to \infty$. In particular this sequence is bounded so we have  a dominating function of the type constant  times $|f(x)|$.
You should also note that $|f(x)-f(1)| <\epsilon$ implies $f(x) <f(1)+\epsilon$ and $f(x) >f(1)-\epsilon$. You can get $\lim \sup n\int f(x)x^{n}dx \leq f(1)$ and $\lim \inf n\int f(x)x^{n}dx \geq f(1)$ to finish the proof.
