This is an question from Mathematical Analysis, Douglass 6.69
Suppose that $f$ is continuous on$[0, 1]$. For each $k \in \mathbb{N}$ and for $x \in [0, 1]$, define $g_k(x)=f(x^k)$. Prove that $$\lim_{k \to \infty} \int_{0}^{1}g_k(x)=f(0)$$
I know that we can exchange the limit and the integral for uniform convergence cases. However, in this case, $g_k$ converges to $f(0)$ if $0\leq x<1$, and $f(1)$ if $x=1$ point wise so this is not a uniform convergence since if $f(0)\neq f(1)$, then the limit is not continuous. I tried to solve the integral just by itself, and taking a limit in the final by substituting $t=x^{1/k}$, but that did not give me any useful information. Can anyone give me some hints?
Hint: Let $\epsilon >0$ and suppose $M=\sup_{0 \leq x \leq 1} |f(x)|$. If $\delta=\frac{\epsilon} M$ then $|\int_{1-\delta}^{1}g_k(x)dx| \leq \epsilon$ since $|g_k(x)| \leq M$. For the integral from $0$ to $1-\delta$ use uniform convergence. [I can add more details if needed].