real analysis - Evaluate $lim_{nrightarrowinfty} int_0^1 f(x^n) dx$

Let $f$ be a real, continuous function defined for all $0\leq x\leq 1$ such that $f(0)=1$, $f(1/2)=2$, and $f(1)=3$. Show that

$$\lim_{n\rightarrow\infty} \int_0^1 f(x^n) dx$$

exists and compute the limit.

Attempt:

Since $f$ is real-valued continuous on $0\leq x\leq 1$ and the boundaries $f(0)=1$ and $f(1)=3$, $f$ is bounded on the interval and the integral $\displaystyle\int_0^1 f(x^n)dx$ exists for any positive $n$. Thus, we can interchange the limit and the integral and compose the limit,
$$\begin{align*} \lim_{n\rightarrow\infty} \int_0^1 f(x^n)dx&=\int_0^1\lim_{n\rightarrow\infty}f(x^n)dx\\ &=\int_0^1 f\left(\lim_{n\rightarrow\infty} x^n \right)dx\\ &=\int_0^1 f(0)dx=1. \end{align*}$$

Questions: So, my first question concerns whether interchanging the limit and the integral is correct. It seems it would be justified by the dominated convergence theorem, where $f(x^n)$ is dominated by a function $g(x)=\displaystyle\sup_{0\leq x\leq 1} f(x^n)$ and $f_n(x^n)$, on $0\leq x\leq 1$, converges pointwise to a function $f$ that takes value 1 for $x\neq 1$ and 3 for $x=1$.

My next question is whether the interchange between the limit and composition of the function is allowed. Since the function is continuous and the limit exists at that point, it seems the interchange would be justified.

Answer

Let $f_n(x) = f(x^n)$ note that the $f_n$ are uniformly bounded (since $f$ is countinuous on $[0,1]$)
and $f_n(x) \to 1$ for all $x <1$.
Then $\lim_n \int f_n = \int \lim_n f_n = 1$.