How to compute the limit below by L'Hopital's rule? 
Let $\left(X_j\right)_{j\geq1}$ be i.i.d. with $E\{X_j\}=\mu$ and $\text{Var}\{X_j\}=\sigma^2$ (all $j$) with $0<\sigma^2<\infty$. Let $S_n=\sum\limits_{j=1}^{n}X_j$ and $Y_n=\dfrac{S_n-n\mu}{\sigma\sqrt{n}}$. Let $\varphi_j$ be the characteristic function of $X_j-\mu$. Since the $\left(X_j\right)_{j\geq1}$ are i.i.d., $\varphi_j$ does not depend on $j$ and we write $\varphi$.
  One can show that $\varphi_{Y_n}(u)=\left(\varphi\left(\dfrac{u}{\sigma\sqrt{n}}\right)\right)^n$. Then, one can expand $\varphi$ in a Taylor expansion about $u=0$ to get
  $$\varphi(u)=1+0-\dfrac{\sigma^2u^2}{2}+u^2h(u)$$ with $h$ denoting the Peano remainder and $h(u)\rightarrow0$ as $u\rightarrow0$.
  One can also show that
  $$\varphi_{Y_n}(u)=e^{n\log\left(1-\frac{u^2}{2n}+\frac{u^2}{n\sigma^2}h\left(\frac{u}{\sigma\sqrt{n}}\right)\right)}$$
  Taking limits as $n\rightarrow\infty$ and using for example L'Hopital rule, one gets $$\lim\limits_{n\to\infty}\varphi_{Y_n}(u)=e^{-\frac{u^2}{2}}$$


MY QUESTION: how could the above limit be solved by using L'Hopital's rule?
This is an indeterminate form $\infty\times0$. Thanks to the advice of user Mark Viola I get to the fact that limit can be rephrased as follows
$$\lim\limits_{n\to\infty}\varphi_{Y_n}(u)=\lim\limits_{n\to\infty}e^{\frac{\log\left(1-\frac{u^2}{2n}+\frac{u^2}{n\sigma^2}h\left(\frac{u}{\sigma\sqrt{n}}\right)\right)}{\frac{1}{n}}}$$
At this point, we get an indeterminate form $\left[\dfrac{0}{0}\right]$. By continuity of exponential function $e$, we have
$$\lim\limits_{n\to\infty}e^{\frac{\log\left(1-\frac{u^2}{2n}+\frac{u^2}{n\sigma^2}h\left(\frac{u}{\sigma\sqrt{n}}\right)\right)}{\frac{1}{n}}}=e^{\lim\limits_{n\to\infty}\frac{\log\left(1-\frac{u^2}{2n}+\frac{u^2}{n\sigma^2}h\left(\frac{u}{\sigma\sqrt{n}}\right)\right)}{\frac{1}{n}}}$$
At this point, focusing on the limit 
$$\lim\limits_{n\to\infty}\frac{\log\left(1-\frac{u^2}{2n}+\frac{u^2}{n\sigma^2}h\left(\frac{u}{\sigma\sqrt{n}}\right)\right)}{\frac{1}{n}}$$
I have tried to solve it by means of L'Hopital rule.
$$\lim\limits_{n\to\infty}\frac{\log\left(1-\frac{u^2}{2n}+\frac{u^2}{n\sigma^2}h\left(\frac{u}{\sigma\sqrt{n}}\right)\right)}{\frac{1}{n}}\stackrel{H}=\lim\limits_{n\to\infty}\frac{\frac{d\log\left(1-\frac{u^2}{2n}+\frac{u^2}{n\sigma^2}h\left(\frac{u}{\sigma\sqrt{n}}\right)\right)}{dn}}{\frac{d\frac{1}{n}}{dn}}$$
I have some problem in computing $$\frac{d\log\left(1-\frac{u^2}{2n}+\frac{u^2}{n\sigma^2}h\left(\frac{u}{\sigma\sqrt{n}}\right)\right)}{dn}$$
Let $f(n)=\left(1-\frac{u^2}{2n}+\frac{u^2}{n\sigma^2}h\left(\frac{u}{\sigma\sqrt{n}}\right)\right)$. I know that $\dfrac{d\log\left(f(n)\right)}{dn}=\dfrac{f^{'}(n)}{f(n)}$ and my problems are related to the computation of $f^{'}(n)$ since I have no clue on how to compute the derivative of $\frac{u^2}{n\sigma^2}h\left(\frac{u}{\sigma\sqrt{n}}\right)$ with respect to $n$, that is
$$\dfrac{d \frac{u^2}{n\sigma^2}h\left(\frac{u}{\sigma\sqrt{n}}\right)}{dn} \tag{1}$$

Could you please explain to me in detail how to solve derivative $(1)$ so as to get to the final result $\lim\limits_{n\to\infty}\varphi_{Y_n}(u)=e^{-\frac{u^2}{2}}$?

HINT:
Let $x=1/n$.  Then examine the limit 
$$\lim_{x\to0}\,\frac{\log\left(1-u^2x/2+(u^2/\sigma^2xh(u\sqrt{x}/\sigma))\right)}{x}$$
Can you proceed now?
