Q:

Convergence of a random variable

I'm working on a problem and I appreciate if you can guide me how to proceed.
Assume that $\sqrt{n}\big(X_n - \mu\big)\rightarrow^d N(0,1) \text{ as } n\rightarrow \infty$. 
By the symbol $\rightarrow^d$ we mean convergence in distribution. 
I want to show that :
$$ \sqrt{n} \big(e^{X_n - \mu} - 1\big) - \sqrt{n}\big(X_n - \mu\big) = o_p(1)$$
I appreciate if you can help me on that.

A:

Note that, for any $z$,
$$
\left|\sqrt{n}\,\left(\exp(z/\sqrt{n})-1\right)-z\right|\leqslant\exp(|z|/\sqrt{n})\,z^2/\sqrt{n}.
$$
Introducing $Z_n=\sqrt{n}(X_n - \mu)$, one sees that
$$
\left|\sqrt{n} \big(\mathrm e^{X_n - \mu} - 1\big) - \sqrt{n}\big(X_n - \mu\big) \right|
\leqslant\mathrm e^{U_n}V_n,
$$
where
$$ 
U_n=|Z_n|/\sqrt{n},\qquad V_n=Z_n^2/\sqrt{n}. 
$$
Since $Z_n$ converges in distribution, $U_n$ and $V_n$ both converge in distribution to $0$, hence $U_n$ and $V_n$ both converge in probability to $0$ hence $\mathrm e^{U_n}V_n$ converges in probability to $0$.
Some general results used above:

If $X_n\to X$ in distribution and $x_n\to0$ then $x_nX_n\to0$ in distribution.
If $X_n\to c$ in distribution, then $X_n\to c$ in probability.
If $X_n\to a$ and $Y_n\to b$ in distribution/probability, then $(X_n,Y_n)\to (a,b)$ in distribution/probability.
If $X_n\to c$ in distribution/probability and the function $u$ is continuous at $c$, then $u(X_n)\to u(c)$ in distribution/probability.

