L'Hospital's Rule Question. show that if $x $ is an element of $\mathbb R$ then $$\lim_{n\to\infty} \left(1 + \frac xn\right)^n = e^x $$
(HINT: Take logs and use L'Hospital's Rule)
i'm not too sure how to go about answer this or putting it in the form $\frac{f'(x)}{g'(x)}$ in order to apply L'Hospitals Rule. 
so far i've simply taken logs and brought the power in front leaving me with 
$$ n\log \left(1+ \frac xn\right) = x $$

Write the limit in the following form $$\lim_{n\to\infty}\frac{\log(1+x/n)}{1/n}=\lim_{n\to\infty}\frac{f(n)}{g(n)}$$ where $f(n)=\log (1+x/n)$ and $g(n)=1/n$ and the limit is in $0/0$ form, so applying L'Hospitals rule we have the above limit is same as $$\lim_{n\to\infty}\frac{f^\prime(n)}{g^\prime(n)}=\lim_{n\to\infty}\frac{1/(1+x/n).(-x/n^2)}{-1/n^2}=\lim_{n\to\infty}\frac{x}{1+x/n}=x$$
