L'hopital limit. Problem I trying to show
$$ \lim_{x\to\infty} \left(\frac{x}{x+1} \right)^x $$
I know the answer using wolfram or a calculator,  but I stuck trying to figure out how to apply L'hopital.

As $x$ approaches infinity:
$$\biggl(\frac{x}{x+1}\biggl)^x=\biggl(1-\frac{1}{x+1}\biggl)^x=e^{\biggl(-\frac{1}{x+1}\biggl)x}=e^{-1}$$ by using Euler's method.
