What is the integration of $\lim_{\sigma\to0}\int\mathcal{N}(x|0,1)f(x_0+x\sigma)dx$? What is the integration of $\lim_{\sigma\to0}\int\mathcal{N}(x|0,1)f(x_0+x\sigma)dx$? Here $\mathcal{N}(x|0,1)$ is a normal Gaussian distribution and $f(\cdot)$ is a continuous and bounded function (i.e. $|f(x)|<C$).
Intuitively speaking, the limit should be $f(x_0)$ because $x$ should be very large in order to make $|f(x_0+x\sigma)-f(x_0)|$ relative large, yet the distance between $f(x_0+x\sigma)$ and $f(x_0)$ is still bounded. In addition, $\mathcal{N}(x|0,1)$ becomes very small when $x$ is very large. So I guess the limit should be $f(x_0)$. 
I tried to use Taylor expansion to prove this. But it requires that the derivatives of any order should exist. And we have to prove that
$$
\lim_{\sigma\to0}\sum_{n=1}^\infty\int\mathcal{N}(x|0,1)f^{(n)}(x_0)\frac{(\sigma x)^n}{n!}dx=0
$$
where $f^{(n)}(x_0)$ is the $n$-th order derivative. Though I can prove for every specific $n$, the limit $\lim_{\sigma\to0}\int\mathcal{N}(x|0,1)f^{(n)}(x_0)\frac{(\sigma x)^n}{n!}dx=0$ if $f^{(n)}(x_0)$ exists, but I  am not sure if the summation also converges to $0$.
How can I prove this rigorously? Thanks.
 A: Thanks for Ian's comment. I studied the Lebesgue's dominant convergence theorem and came up with the following proof.
Let $\{f_n(x)\}$ be a sequence of real valued functions defined as
$$
f_n(x)=\mathcal{N}(x|0,1)f(x_0+\frac{x}{n})
$$
So $\{f_n(x)\}$ converges to $\mathcal{N}(x|0,1)f(x_0)$ pointwise. Moreover, since $f(x)$ is bounded, i.e. $|f(x)|\le C$, so we have
$$
|f_n(x)|\le\mathcal{N}(x|0,1)C
$$
for all $n$. As $\mathcal{N}(x|0,1)C$ is integrable, so according to Lebesgue's dominant convergence theorem, the limit of $\{f_n(x)\}$, i.e. $\mathcal{N}(x|0,1)f(x_0)$, is also integrable and
$$
\lim_{n\to\infty}\int f_n(x)dx=\int f(x)dx=\int\mathcal{N}(x|0,1)f(x_0)dx=f(x_0)
$$
So we have
$$
\lim_{\sigma\to0}\int\mathcal{N}(x|0,1)f(x_0+x\sigma)dx=f(x_0)
$$
The Lebesgue's dominant convergence theorem is stated as

Let $\{f_n\}$ be a sequence of real-valued measurable functions on a measure space $(S,\Sigma,\mu)$. Suppose that the sequence converges point-wise to a function $f$ and is dominated by some integrable function $g$ in the scense that
  $$
|f_n(x)|\le g(x)
$$
  for all numbers $n$ in the index set of the sequence and all points $x\in S$. Then $f$ is integrable and 
  $$
\lim_{n\to\infty}\int_Sf_n(x)d\mu=\int_Sf(x)d\mu
$$
  (https://en.wikipedia.org/wiki/Dominated_convergence_theorem)

In our case, $S$ is $\mathbb{R}$, $\mu$ is the Lebesgue measure. $g(x)$ is $\mathcal{N}(x|0,1)C$.
