In the proof of the delta method (about little oh_p) In the proof of the delta method related with the convergence in distribution, I couldn't understand the statement below.
When $ \sqrt{n} (X_n - \mu) \rightarrow ^D N(0, \sigma^2 ) $ ,
\begin{equation}
f(X_n) = f(\mu) + f'(\mu)(X_n -\mu ) + o_p (|X_n - \mu |),
\end{equation}
where $a_n = o_p(b_n)$ means that $ \frac{a_n}{b_n} \rightarrow^p  0$  as $n \rightarrow  \infty $.
I think I have to show this statement is true.
\begin{equation}
g(X_n) \equiv \frac{f(X_n) - f(\mu)}{X_n -\mu} -  f'(\mu) \rightarrow^p 0  
\end{equation}
In some textbooks and pdf files, they define a new countuous function h(x) such that 
$$
h(x) = 
\begin{cases}
g(x) & \text{if x $\ne$ 0}\\
0 & \text{if x $\eqcirc$ 0}\
\end{cases}
$$ 
Then, by continuous mapping theorem, $ h(X_n) \rightarrow ^p 0$.
So, they say $g(X_n) \rightarrow ^p 0$.
But, because $ h(X_n) \ne g(X_n)$ (at x=0 ), I couldn't convince myself that $ h(X_n) \rightarrow ^p 0$ => $g(X_n) \rightarrow ^p 0$ . 
Is there no problem of insisting that $ h(X_n) \rightarrow ^p 0$ => $g(X_n) \rightarrow ^p 0$  ?
Thanks for reading my question in advance.
 A: First of all I think $a = \mu$ in the statement of your question. Moreover you have to use the fact that $X_n \overset{P}{\to} \mu$ (this fact can be derived from $\sqrt{n} (X_n - \mu) \overset{D}{\to} N(0, \sigma^2 )$ but I think you have derived it before). 
The function $$g(x) \equiv \frac{f(x) - f(\mu)}{x -\mu} -  f'(\mu)$$ is not defined at $x=\mu$ but it admits a continuous extension at $x=\mu$. To show that, one has to check that the function $$ h(x) = 
\begin{cases}
g(x) & \text{if } x \ne \mu\\
0 & \text{if } x = \mu\
\end{cases}$$
is continuous at $x=\mu$. This function $h$ is the continuous extension  of $g$  at $x=\mu$. By the continuous mapping theorem and knowing that  $X_n \overset{P}{\to} \mu$, one has then $h(X_n) \overset{P}{\to} h(\mu)=0$. But since $h$ is the continuous extension  of $g$  at $x=\mu$, then this limit also holds for $g$, that is, one has $g(X_n) \overset{P}{\to} 0$. This latter fact is well known in elementary real analysis for classical limits, but it also holds true for limits in probability (proving it is a good elementary exercise about convergence in probability).
