# 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!

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 $$\lim_{t\to +\infty}\left(1+\frac{1}{t}\right)^{t}=e.$$

• Thanks a lot! This really helped. Commented Sep 3, 2017 at 11:07
• But can your clarify how you simplified the expression. I am having a hard time understanding how you turned $\dfrac{x^x}{\left(x+2\right)^x}$ into $\dfrac{1}{\left(\frac{2}{x}+1\right)^x}$ @Robert Z Commented Sep 3, 2017 at 13:29
• @Abu Bakr Is it better now? Commented Sep 3, 2017 at 13:51
• Yes, thank you. Commented Sep 3, 2017 at 14:00

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...

HINT: write $$\frac{1}{\left(\left(1+\frac{2}{x}\right)^{2x}\right)^{1/2}}$$

• What is the limit of $(1+\frac 2x)^{2x}$? It seems to me that you wanted to write what Robert Z has written in his answer. Commented Sep 3, 2017 at 11:13
• this Limit is $e^4$ Commented Sep 3, 2017 at 11:17
• Thank you for ignoring pragmatic implication. Commented Sep 3, 2017 at 11:20

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:

$$\lim_{x\rightarrow+\infty}\frac{x^x}{(x+2)^x}=\lim_{x\rightarrow+\infty}\left(1+\frac{x}{x+2}-1\right)^{\frac{1}{\frac{x}{x+2}-1}\cdot\left(-\frac{2x}{x+2}\right)}=e^{-\lim\limits_{x\rightarrow+\infty}\frac{2x}{x+2}}=e^{-2}=\frac{1}{e^2}.$$

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}

• Using the Taylor series at $x=0$ and then setting $x\to\infty$ requires a high degree of faith... even if it works :) Commented Sep 3, 2017 at 11:25
• @Miguel, when $x$ is nearing infinity, $-2/(x+2)$ is sufficiently close to zero, why wouldn't Taylor work? I'll see if I can rewrite it in more transparent way. Commented Sep 3, 2017 at 11:29
• I get why it works, it is just that proving formally that the series converge at $x\to\infty$ is well beyond Calculus I. Commented Sep 3, 2017 at 11:38
• @Miguel, is my edit satisfactory? Commented Sep 3, 2017 at 12:02