Let $f: [0 ,1] \to \mathbb{R} $ be continuous, prove $\lim_{n\to \infty} \int_0^1 f(x^n)dx = f(0)$
As I mentioned in a comment, if you allow the Dominated Convergence Theorem then this is a matter of a few lines. The sequence of functions $(f_n)_n$ defined by $f_n(x) \stackrel{\rm def}{=} f(x^n)$ ($x\in[0,1])$ converges pointwise (by continuity of $f$) to the constant function $f(0)$ on $[0,1)$, and one can uniformly dominate all $f_n$'s by the integrable (because constant) function $g\stackrel{\rm def}{=} \lVert f\rVert_\infty$ ($f$ being bounded on $[0,1]$, since it is continuous). Now, by the DCT we get $$ \int_0^1 f_n \xrightarrow[n\to\infty]{} \int_0^1 \lim_{n\to\infty}f_n = \int_0^1 f(0)dx = f(0). $$
But this is overkill. Another option, explored below, would be to use continuity of $f$ at $0$: recall that $f$ being continuous on a closed interval, it is bounded.
Without loss of generality, we can assume $f(0) = 0$ (indeed, just apply the same exercise to $g\stackrel{\rm def}{=} f-f(0)$. It is easy to check that this will be sufficient to prove the result).
Fix $\varepsilon>0$, break the interval in 2: $$ \int_0^1 f(x^n)dx = \int_0^a f(x^n)dx +\int_a^1 f(x^n)dx $$ for some convenient $a=a(\varepsilon, f) \stackrel{\rm def}{= } 1-\frac{\varepsilon}{2\lVert f\rVert_\infty}$. The second part has absolute value at most $(1-a)\cdot\lVert f\rVert_\infty < \frac{\varepsilon}{2}$ by definition.
As for the first, we use continuity of $f$ at $0$: let $\delta_\varepsilon>0$ be such that $\lvert u\rvert \leq \delta_\varepsilon$ implies $\lvert f(u)\rvert \leq \frac{\varepsilon}{2}$. We then rely on the fact that for $n$ big enough (i.e., for $n\geq N_{\varepsilon}$, for some $N_{\varepsilon}\geq 0$, we have $\lvert x^n\rvert \leq \lvert a^n\rvert \leq \delta_\varepsilon$. Thus, the first term can be bounded as $$ \left\lvert \int_0^a f(x^n)dx\right\rvert \leq \int_0^a \left\lvert f(x^n)\right\rvert dx \leq \int_0^a \frac{\varepsilon}{2} dx \leq \frac{\varepsilon}{2} $$ for any $n\geq N_{\varepsilon}$.
Putting it together, for $n\geq N_{\varepsilon}$ one has $$ \left\lvert \int_0^1 f(x^n)dx\right\rvert \leq \frac{\varepsilon}{2} + \frac{\varepsilon}{2} = \varepsilon $$ by the triangle inequality. This shows that $$ \int_0^1 f(x^n)dx\xrightarrow[n\to\infty]{} 0 $$ and from there the general case follows.
Since $f $ is continuous, given $\varepsilon>0$ there exists $\delta>0$ such that $|f (x)-f (0)|<\varepsilon/2 $ whenever $|x|<\delta $. Let $M=\max \{|f (x)|:\ x\in [0,1]\}$. Choose $x_0\in (0,1) $ such that $1-x_0 <\varepsilon/2M $. Choose $n_0$ such that $|x_0^{n_0}|<\delta $ . Then, for any $n\geq n_0$,
\begin{align} \left|\int_0^1f (x^n)\,dx-f (0)\right| &=\left|\int_0^1(f (x^n)-f (0))\,dx\right| \leq \int_0^1|f (x^n)-f (0)|\,dx \\ \ \\ &= \int_0^{x_0}|f (x^n)-f (0)|\,dx + \int_{x_0}^1|f (x^n)-f (0)|\,dx \\ \ \\ &\leq \frac\varepsilon2 x_0+2M (1-x_0)\\ \ \\ &<\frac\varepsilon 2+\frac\varepsilon 2=\varepsilon. \end{align}
As an aside, if you have the Fundamental Theorem of Calculus available, the substitution $u=x^n $ does the job.