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 $L$.