Show that $\lim_{n\to\infty}\int_0^1f(x^n)dx = f(0)$ Given a continuous function $f:[0,1] \to R$, prove that $\lim_{n\to\infty}\int_0^1f(x^n)dx = f(0)$.
Attempt:
Let $u = x^n$, so $du = nx^{n-1}dx$. Then substituing $u$ in I got:
$\lim_{n\to\infty}\dfrac{\int_0^1f(u)du}{nx^{n-1}}$
Doesn't this limit go to $0$? I'm not sure which part I'm messing up on, any hints are appreciated.

It's equivalent to show $\lim_{n\rightarrow\infty}\int_0^1 (f(x^n)-f(0))dx=0$.
Given any $\epsilon>0$, we have $$\left|\int_0^1 (f(x^n)-f(0))dx\right|\le\int_0^1 |f(x^n)-f(0)|dx$$ $$=\int_0^{1-\epsilon}|f(x^n)-f(0)|dx + \int_{1-\epsilon}^1 |f(x^n)-f(0)|dx$$ $$\le\int_0^{1-\epsilon}|f(x^n)-f(0)|dx + 2B\epsilon$$
where $B=\max_{x\in[0,1]}|f(x)|$ which exists due to continuity.
By $f(x)$ is continuous at $x=0$, there is a $\delta>0$, such that $|x|\le\delta\Rightarrow |f(x)-f(0)|\le \epsilon$. Thus when $n\ge \log_{1-\epsilon}\delta$, $|x^n|\le (1-\epsilon)^n\le\delta$ for $x\in [0, 1-\epsilon]$, and for sufficiently large $n$, $$\int_0^{1-\epsilon} |f(x^n)-f(0)|dx\le \int_0^1 \epsilon dx \le \epsilon$$
Finally,
$$\left|\int_0^1 (f(x^n)-f(0))dx\right|\le (1+2B)\epsilon$$
In fact, we only need $f(x)$ is bounded and right continuous at $0$ (and integrable).
