Find $\lim \limits_{n \to \infty} \int_0^1 f(x) g(x^n) dx$ Find   $$\lim \limits_{n \to \infty} \int_0^1 f(x) g(x^n) dx$$
Where 
$\;f\in C^0, \quad g\in D^1$  
Any hints? 
 A: You don't really need $g$ to be differentiable, it is enough to assume $g \in C[0,1]$ (in fact, all you need is to have $g$ integrable and bounded, and continuous at $0$, as the proof below shows). Then the limit in question equals 
$$
g(0) \int_0^1 f(x) dx,
$$
which we prove below.
Fix $\delta>0$ small and split the integral into $2$ parts, namely
$$
\int_0^1 f(x) g(x^n) dx = \int_0^{1-\delta} + \int_{1-\delta}^1 := I_1 + I_2.
$$
In view of continuity of $f$ and $g$, both functions are bounded on $[0,1]$ and we get
\begin{equation}
|I_2| \leq \delta ||f||_{\infty} ||g||_\infty. \tag{1}
\end{equation}
For $I_1$, observe that $x\in [0,1-\delta]$, hence for any $\varepsilon>0$ small by choosing $N\in \mathbb{N}$ large enough we get
\begin{equation}
|g(x^n) -g(0)| \leq \varepsilon, \ \ \forall x\in[0,1-\delta] \text{ and } \forall n>N, 
\end{equation}
which follows by continuity of $g$ at $0$.
From here, for all $n>N$ we obtain
\begin{equation}
\left| I_1 - g(0) \int_{0}^{1-\delta} f(x) dx \right|  \leq \int_0^{1-\delta} |f(x)| |g(x^n) - g(0)| dx \leq \varepsilon \int_0^1|f(x)| dx. \tag{2}
\end{equation}
Since $\delta>0$ and $\varepsilon>0$ are arbitrary small numbers, from $(1)$ and $(2)$ we get the claim.
A: $f(x)g(x^{n}) \to f(x) g(0)$ almost everywhere and it is uniformly bounded, so teh limit is $g(0)\int_0^{1} f(x) \, dx$ by Bounded Convergence Theorem.
A: Solution
We only need to assume that:

*

*$f$ is integrable over $[0,1]$;

*$g$ is bounded over $[0,1]$ and continuous at $x=0$.

Thus, denote $\phi_n(x)=f(x)g(x^n)$ where $x \in [0,1],n=1,2,\cdots$, and $\max\limits_{0\leq x \leq 1} |g(x)|=M$. It's easy to have that
$$|\phi_n(x)| \leq M\cdot|f(x)|,$$ for $n=1,2,\cdots.$ Moreover, the dominating function $M \cdot|f(x)|$ is obviously integrale over $[0,1].$ Besides, $$\lim_{n \to \infty}\phi_n(x)=f(x)g(0),~~~~x \in [0,1].$$
As a result, by Lebesgue's dominated convergence theorem, we have $$\lim_{n \to \infty}\int_0^1 \phi_n(x){\rm d}x=\int_0^1 f(x)g(0){\rm d}x=g(0)\int_0^1f(x){\rm d}x.$$
