Ytukyg

Suppose that $f:[0,1]tomathbb{R}$ is continuous on $[0,1]$. Show that ${int_0^1f(x^n)dx}_{n=1}^infty$ converges to $f(0)$.

I'm not really sure to go about proving this. I know $lim_{ntoinfty}x^n=0$ for $x<1$ so it makes sense that the sequence would converge to $int_0^1f(0)dx=f(0)$. I have no idea how to formalize this and make it into a proof though. I'm really bad at this so I'd really appreciate a thorough explanation.

Since $f(x)$ is continuous, for any $xin(0,1)$ the sequence $f(x^n)$ is convergent to $f(0)$ as $nto +infty$.
On the other hand the continuity of $f$ on $[0,1]$ implies its boundedness, hence the dominated convergence theorem trivially applies.

Hint

Let $epsilon>0$ small enough.

$$int_0^{1-epsilon}(f(x^n)-f(0))dx=(1-epsilon)(f(c^n)-f(0))$$
with $0le cle 1-epsilon<1$.
now use sequential charactersation of the continuity at $0$.

As $c^nto 0$, For large $n$,

$$|int_0^{1-epsilon}(f(x^n)-f(0))dx|<frac{epsilon}{2}.$$
On the other hand,
$$|int_{1-epsilon}^1(f(x^n)-f(0))dx|le 2Mepsilon.$$
with $M=sup_{[0,1]}|f|$.