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.

$\newcommand{\d}{\ {\mathrm d}} \newcommand{\e}{\varepsilon}\newcommand{\md}[1]{\left\lvert #1 \right\rvert}$Here's a bare-hands elementary proof without any appeal to the Dominated Convergence Theorem.
Note that we have $$\int_0^1 f(x^n) \d x - f(0) = \int_0^1 [f(x^n) - f(0)] \d x.$$
Let $M := \sup_{[0, 1]} |f|$. Note that this is finite since $f$ is continuous.
Let $\e > 0$ be given. Fix $\eta \in (0, 1)$ such that $2 (1 - \eta) M < \frac{\e}{2}$.
By continuity at $0$, there exists $\delta > 0$ such that $$|y| < \delta \Rightarrow |f(y) - f(0)| < \frac{\e}{2}.$$
Note that $x \mapsto \md{x^n}$ is increasing on $[0, \eta]$ and $\eta^n \to 0$. Thus, there exists $N \in \Bbb N$ such that $$|x^n| < \delta \text{ for all } x \in [0, \eta] \text{ and all } n \geqslant N.$$
Thus, for $n \geqslant N$, we have
\begin{align}
\md{\int_0^1 [f(x^n) - f(0)] \d x} & \leqslant \int_0^1 \md{f(x^n) - f(0)} \d x \\
&= \int_0^{\eta} \md{f(x^n) - f(0)} \d x + \int_{\eta}^{1} \md{f(x^n) - f(0)} \d x \\
&\leqslant \int_0^{\eta} \frac\e2 \d x + \int_{\eta}^{1} 2M \d x \\
&\leqslant \frac\e2 \eta + 2M (1 - \eta) \\
&< \e.
\end{align}
