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
$$ n\log \left(1+ \frac xn\right) = x $$
Best Answer
The ‘$=x$’ is getting ahead of yourself a bit. Let $$L=\lim_{n\to\infty}\left(1+\frac{x}n\right)^n\;,$$ and take the logarithm to get
$$\begin{align*} \ln L&=\ln\lim_{n\to\infty}\left(1+\frac{x}n\right)^n\\ &=\lim_{n\to\infty}\ln\left(1+\frac{x}n\right)^n\\ &=\lim_{n\to\infty}n\ln\left(1+\frac{x}n\right)\;, \end{align*}$$
where the interchange of the log and the limit is justified by the fact that the logarithm function is continuous.
This limit is now a so-called $\infty\cdot 0$ indeterminate form, and there is a standard approach to those: move one of the factors into the denominator. In this case we have
$$\ln L=\lim_{n\to\infty}\frac{\ln\left(1+\frac{x}n\right)}{1/n}\;,$$
a limit in which both numerator and denominator approach $0$ as $n\to\infty$. Now you can apply l’Hospital’s rule.
Don’t forget that at this point you’re actually finding $\ln L$, not $L$, so you’ll have to exponentiate to get $L$.