Evaluate limit of $\lim_{x \to \infty}\frac{x^x}{\left(x+2\right)^x}$ $\lim_{x \to \infty}\dfrac{x^x}{\left(x+2\right)^x}$
I tried using Taylor and L'H and wasn't able to land on an answer.
Any help would be appreciated!

An approach with Taylor series. Let $f(x) = \frac{x^x}{(x+2)^x}$. Then,
\begin{align}
\ln f(x) = x\ln\left(1-\frac 2{x+2}\right)=x\left(-\frac{2}{x+2}-\frac 4{(x+2)^2}+\ldots\right)\stackrel{x\to +\infty}{\to}-2
\end{align}
which implies $\lim_{x\to+\infty}f(x) = e^{-2}$.
EDIT:
As Miguel points out, it might be problematic to use the above technique, because what we really are using is not Taylor series, but Laurent series.
To avoid the issue, we could do as follows:
\begin{align}
\lim_{x\to +\infty} x\ln\left( 1-\frac 2{x+2} \right) &= \lim_{t\to 0^+}\frac 2 t\ln\left(1 -\frac 2{\frac 2t + 2} \right)\\
&= \lim_{t\to 0^+}\frac 2 t\ln\left(1 -\frac {2t}{2 + 2t} \right)\\
&= \lim_{t\to 0^+}\frac 2 t\ln\frac {1}{1 + t}\\
&= -2\lim_{t\to 0^+}\frac{\ln(1+t)}t = -2
\end{align}
