delta-method-ish question This is a problem from Resnick's heavy tail analysis book:
Let $\{X_n\}$ be a sequence of random variables such that $EX_n=m$ and Var$(X_n)=\sigma_n^2>0$ for all $n$, where $\sigma_n^2\rightarrow 0$ as $n\rightarrow\infty$ Define $$Z_n=\sigma_{n}^{-1} (X_n-m)$$ and let $f$ be a function with non-zero derivative $f'(m)$ at $m$. Show that 
(1) $X_n-m\Rightarrow0$
(2)   If $$Y_n=\frac{f(X_n)-f(m)}{\sigma_n^{-1}f'(m)}$$show that $Y_n-Z_n\Rightarrow 0$ 
(3)Show that if $Z_n$ converges in probability or in distribution, then so does $Y_n$.
(4)If $S_n$ is binomially distributed with parameters $n$ and $p$ and $f'(p)\ne0$, use the preceding results to determine the asymptotic distribution of $f(S_n/n)$
Attempted solution
(1)It suffices to show that $X_m-m\rightarrow 0$ in probability which can follows easily by Chebyshev's inequality.
(2) I am not sure how to do this one. Since we are only given that the function is differentiable at $m$, it seems that Taylor theorem is the way to go. So rewriting this question using Taylor expansion in the first degree, we have $$f(X_n)=f(m)+f'(m)(X_n-m)+h(X_n)(X_n-m)$$ and rearrange the term and dividing by $\sigma_n$ we get $$\frac{f(X_n)-f(m)}{\sigma_n f'(m)}-\frac{X_n-m}{\sigma_nf'(m)}=\frac{h(X_n)(X_n-m)}{\sigma_n f'(m)}$$ I got stuck here since I don't know what to do with RHS..
I didn't figure out (3) and (4) since i believe they rely on the result of (2)..Can someone explain what this problem is trying to say? I think it looks like Delta method in statistics..
 A: Here is a proof for $(2)$ assuming that $f'(x)$ is continuous at $x=m$(we can get rid of this assumption without particular difficulty, see the end of answer)
\begin{align}
Y_n - Z_n & = \frac{f(X_n)-f(m)}{\sigma_n^{-1}f'(m)} - \frac{X_n - m}{\sigma_n} \\
& = \frac{f'(\xi_n)(X_n - m)}{\sigma_n^{-1}f'(m)} - \frac{X_n - m}{\sigma_n} \\
& = \left(\frac{f'(\xi_n)}{f'(m)} - 1\right)\frac{X_n - m}{\sigma_n} \\
\end{align}
where $\xi_n$ lies between $X_n$ and $m$. Since $X_n$ converges to $m$ in probability, so does $\xi_n$. Then the assumption that $f'(x)$ is continuous at $x=m$ gives $\frac{f'(\xi_n)}{f'(m)} - 1$ converges to $0$ in probability.
For any given $\epsilon >0$, we have
\begin{align}
P(|Y_n - Z_n| > \epsilon) \leq & P(\left|\frac{X_n - m}{\sigma_n}\right| > N) + P(\left|\frac{X_n - m}{\sigma_n}\right| \leq N, \left|\frac{f'(\xi_n)}{f'(m)} - 1\right| \geq \frac{\epsilon}{N}) \\
\leq & P(\left|\frac{X_n - m}{\sigma_n}\right| > N) + P(\left|\frac{f'(\xi_n)}{f'(m)} - 1\right| \geq \frac{\epsilon}{N}) \\
\end{align}
and we have that
$$P(\left|\frac{X_n - m}{\sigma_n}\right| > N) \leq \dfrac{E\left(\frac{X_n - m}{\sigma_n}\right)^2}{N^2} =\dfrac{1}{N^2}$$
For any $\delta >0$, take $N$ such that $\dfrac{1}{N^2} < \dfrac{\delta}{2}$, then the convergence in probability of $\frac{f'(\xi_n)}{f'(m)} - 1$ implies when $n$ is large enough, we have $P(\left|\frac{f'(\xi_n)}{f'(m)} - 1\right| \geq \frac{\epsilon}{N}) < \frac{\delta}{2}$
In summary, we have proven $Y_n - Z_n$ converges in probability to $0$
Then $(3)$ follows from Slutsky's theorem
A similar argument using the Peano form of Taylor reminder enables to get rid of the assumption of continuity, i.e. with $OP$'s notation, we have $h(X_n)$ converges to $0$ in probability, then repeat above arguments to show $h(X_n)\dfrac{X_n - m}{\sigma_n f'(m)}$ converges to $0$ in probability
