If $f(0)=1$, $f(1/2)=2$, $f(1)=3$, show that $\lim_{n \to \infty} \int_0^1 f(x^n) \ dx$ exists and compute the limit. 
Let $f(x)$ be continuous on the closed interval $[0,1]$ and satisfy $f(0)=1$, $f(1/2)=2$, $f(1)=3$. Show that $$\lim_{n \to \infty} \int_0^1 f(x^n) \ dx$$ exists and compute its limit.

I plotted the function $2x^n+1$ on desmos and I see that most values assume $1$ as $n \to \infty$. So, my guess is that the integral is going to approach $1$. 
If we also compute $\int_0^1 (2x^n+1) \ dx=\frac{2x^{n+1}}{n+1}+x\Big|_0^1=\frac{2}{n+1}+1$, we see that to $\frac{2}{n+1}+1 \to 1$ as $n \to \infty$.
How can I make this rigorous?
 A: For each $x \in (0,1)$, $x^n \to 0$ as $n \to \infty$, so by continuity, $f(x^n) \to f(0)$. Since $f$ is continuous on $[0,1]$ it is bounded, say by some constant $C$, so that $\forall x,n$, $|f(x^n)| \le C$. Since $C$ is integrable on $[0,1]$, dominated convergence theorem tells us $\int_0^1 f(x^n)dx \to \int_0^1 f(0)dx = 1$.
A: By change of variable we have
\begin{align*}
\int_{0}^{1}f(x^{n})dx=\dfrac{1}{n}\int_{0}^{1} u^{n^{-1}-1}f(u)du,
\end{align*}
and we also know that
\begin{align*}
\int_{0}^{1}u^{n^{-1}-1}du=n,
\end{align*}
so
\begin{align*}
\dfrac{1}{n}\int_{0}^{1}u^{n^{-1}-1}f(u)du-1&=\dfrac{1}{n}\int_{0}^{1}u^{n^{-1}-1}f(u)du-\dfrac{1}{n}\int_{0}^{1}f(0)u^{n^{-1}-1}du\\
&=\dfrac{1}{n}\int_{0}^{1}u^{n^{-1}-1}(f(u)-f(0))du.
\end{align*}
Given $\epsilon>0$, since $f$ is continuous, there exists some $\delta\in(0,1)$ such that for all $u\in(0,\delta)$, we have $|f(u)-f(0)|<\epsilon$. Now,
\begin{align*}
&\dfrac{1}{n}\int_{0}^{1}u^{n^{-1}-1}|f(u)-f(0)|du\\
&=\dfrac{1}{n}\int_{0}^{\delta}u^{n^{-1}-1}|f(u)-f(0)|du+\dfrac{1}{n}\int_{\delta}^{1}u^{n^{-1}-1}|f(u)-f(0)|du\\
&\leq\dfrac{1}{n}\int_{0}^{1}u^{n^{-1}-1}\cdot\epsilon du+\dfrac{2M}{n}\int_{\delta}^{1}u^{n^{-1}-1}du\\
&=\epsilon+\dfrac{2M}{n}\cdot n(1-\delta^{n^{-1}})\\
&=\epsilon+2M(1-\delta^{n^{-1}}),
\end{align*}
where $M:=\sup_{x\in[0,1]}|f(x)|$. We have $\delta^{n^{-1}}\rightarrow 1$ as $n\rightarrow\infty$, so 
\begin{align*}
\dfrac{1}{n}\int_{0}^{1}u^{n^{-1}-1}|f(u)-f(0)|du\rightarrow 0,
\end{align*}
and hence
\begin{align*}
\dfrac{1}{n}\int_{0}^{1}u^{n^{-1}-1}f(u)du\rightarrow 1.
\end{align*}
A: Since $f$ is continuous on a compact set its bounded by some constant function $M$. Infact, choosing the sequence of measurable functions $ f_n(x):= f(x^n)$ we have 
$$ |f_n(x)| \leq M \ \forall x , n $$
Also notice $f_n$ converges to $1$ pointwise for $x \in [0,1)$.
By the dominated convergence theorem
$$ \lim_{n \rightarrow \infty} \int f_n = \int 1 = 1$$
A: Without the dominated convergence theorem:


*

*First, $f$ is continuous, so is bounded on $[0,1]$: Let$C>0$ be such that $\sup_{[0,1]} \lvert f\rvert + 1 \leq C$. Also, $x\mapsto f(x^n)$ is continuous as well, so the integral $\int_0^1 f(x^n)dx$ is well-defined for all $n\in\mathbb{N}$.

*Fix $\varepsilon > 0$, and let $\delta$ be such that (by continuity at $0$) $$\lvert f(x) - 1\rvert \leq \varepsilon$$ for all $x\in[0,1]$ such that $\lvert x\rvert \leq \delta$.

*Consider $n_0\in\mathbb{N}$ such that $\left(1-\frac{\varepsilon}{C}\right)^n \leq \delta$ for all $n\geq n_0$. 
Then, for all $n\geq n_0$, we have $\lvert x^n\rvert \leq \delta$ for all $x\in[0,1-\frac{\varepsilon}{C}]$, and we will use that to handle
$$
\int_0^1 f(x^n)dx = \int_0^{1-\frac{\varepsilon}{C}} f(x^n)dx
+ \int_{1-\frac{\varepsilon}{2C}}^1 f(x^n)dx \tag{1}
$$
For the second term:
$$
\left\lvert \int_{1-\frac{\varepsilon}{2C}}^1 f(x^n)dx\right\rvert \leq 
\int_{1-\frac{\varepsilon}{2C}}^1 \left\lvert f(x^n)\right\rvert dx
\leq \int_{1-\frac{\varepsilon}{2C}}^1 Cdx = \varepsilon\,. \tag{2}
$$
For the first term:
$$
1\leq \left(1-\frac{\varepsilon}{C}\right)\left(1+\varepsilon\right) \leq \int_0^{1-\frac{\varepsilon}{C}} \left(1-\varepsilon\right)dx \leq \int_0^{1-\frac{\varepsilon}{C}} f(x^n)dx\leq \int_0^{1-\frac{\varepsilon}{2C}} \left(1+\varepsilon\right)dx \leq 1+\varepsilon. \tag{3}
$$
using continuity at $0$ and our choice of $n_0$.
Overall, using (1), (2), and (3),
$$
1-2\varepsilon \leq \int_0^1 f(x^n)dx \leq 1+2\varepsilon
$$
for all $n\geq n_0$. This shows the convergence to $1$, as $\varepsilon$ was arbitrary.
