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!
 A: Note that as $x \to \infty$,
$$\frac{x^x}{\left(x+2\right)^x}=\left(\frac{x}{x+2}\right)^{x}=
\left(\frac{1}{1+\frac{2}{x}}\right)^{x}=\left(1+\frac{1}{x/2}\right)^{-x}=\left(\left(1+\frac{1}{x/2}\right)^{x/2}\right)^{-2}\to e^{-2}.$$
where we used the well-known limit
$$\lim_{t\to +\infty}\left(1+\frac{1}{t}\right)^{t}=e.$$
A: You can take logarithms, which has the advantage that you do not have to recall well-known limits.
If 
$$L=\lim_{x\to\infty}\left(\frac{x}{x+2}\right)^x$$
then
$$\ln L=\lim_{x\to\infty}x\,\left(\ln x-\ln(x+2)\right)=\lim_{x\to\infty}\frac{\ln x-\ln(x+2)}{\frac{1}{x}}$$
Can you take it from here?, e.g. L'Hôpital...
A: For $x\rightarrow-\infty$ and for $x\rightarrow\infty$ your limit does not exist because the domain of $x^x$ is $(0,+\infty)$.
By the way, since $e^x$ and $\ln x$ are continuous functions, we obtain:
$$\lim_{x\rightarrow+\infty}\frac{x^x}{(x+2)^x}=\lim_{x\rightarrow+\infty}\left(1+\frac{x}{x+2}-1\right)^{\frac{1}{\frac{x}{x+2}-1}\cdot\left(-\frac{2x}{x+2}\right)}=e^{-\lim\limits_{x\rightarrow+\infty}\frac{2x}{x+2}}=e^{-2}=\frac{1}{e^2}.$$
A: HINT: write $$\frac{1}{\left(\left(1+\frac{2}{x}\right)^{2x}\right)^{1/2}}$$
A: 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}
