Compute $\lim \limits_{x\to\infty} (\frac{x-2}{x+2})^x$ Compute 
$$\lim \limits_{x\to\infty} (\frac{x-2}{x+2})^x$$
I did
$$\lim_{x\to\infty} (\frac{x-2}{x+2})^x = \lim_{x\to\infty} \exp(x\cdot \ln(\frac{x-2}{x+2})) =  \exp( \lim_{x\to\infty} x\cdot \ln(\frac{x-2}{x+2}))$$
But how do I continue? The hint is to use L Hopital's Rule. I tried changing to 
$$\exp(\lim_{x\to\infty} \frac{\ln(x-2)-\ln(x+2)}{1/x})$$
This is 
$$(\infty - \infty )/0 = 0/0$$
But I find that I can keep differentiating? 
 A: Hint :
Rewrite limit into form :
$$\lim_{x\to\infty} \left(1+\frac{1}{\left(\frac{x+2}{-4}\right)}\right)^{\left(\frac{x+2}{-4}\right) \cdot \left(\frac{-4x}{x+2}\right)}$$
A: A nitpick: $\infty-\infty$ is not 0! It's undefined. Your limit is of the form $0/0$ though.
You can apply L'H'ôpital from the start if you like: $\lim\limits_{x\rightarrow\infty}{x-2\over x+2} =1$, and $\ln 1=0$. 
So
$$
 \lim_{x\rightarrow\infty} \Bigl(x \ln{x-2\over x+2} \Bigr)
=\lim_{x\rightarrow\infty}  {\ln{x-2\over x+2}\over1/x}
=\lim_{x\rightarrow\infty}   {{x+2\over x-2}\cdot{1(x+2)-1(x-2)\over (x+2)^2} \over- 1/x^2 }
=\lim_{x\rightarrow\infty}  {{-4x^2\over (x+2) (x-2)}   }=-4.
$$
(use L'Hopital again to evaluate the limit on the right hand side if you like).
So,
$$\lim_{x\rightarrow\infty}\Bigl({x-2\over x+2}\Bigr)^x
=e^{ \lim\limits_{x\rightarrow\infty}\bigl(x\ln{x-2\over x+2}\bigr)}=e^{-4}.
$$

To answer more directly, L'Hôpital applied to $$\lim_{x\rightarrow\infty}{\ln(x-2)-\ln(x+2)\over 1/x}$$ gives you
 $$\lim_{x\rightarrow\infty}{{1\over x-2}-{1\over x+2}\over- 1/x^2}.$$
Now simplify:
$$
{{1\over x-2}-{1\over x+2}\over- 1/x^2}
=-x^2\Bigl({1\over x-2}-{1\over x+2}\Bigr)
= {-4x^2\over (x+2)(x-2)}.
$$ 
So, using L'Hôpital's rule again
$$
\lim_{x\rightarrow\infty}{{1\over x-2}-{1\over x+2}\over- 1/x^2}
=\lim_{x\rightarrow\infty} {-4x^2\over (x+2)(x-2)}
=\lim_{x\rightarrow\infty} {-8x\over (x+2)+(x-2)}
=\lim_{x\rightarrow\infty} {-8x\over2x}=-4.
$$
A: This can be done using only the definition of $e$, 
$$e = \lim_{n\to\infty}(1+1/n)^n.$$ 
Notice that this implies immediately that $1/e =  \lim_{n\to\infty}(1-1/n)^n$ 
and, more generally, 
$$\lim_{n\to\infty} (1+ a/n)^n = e^{a n}.$$ 
We find
$$\lim_{x\to\infty} \left(\frac{x-2}{x+2}\right)^x
= \lim_{x\to\infty} \left(\frac{1-2/x}{1+2/x}\right)^x
= \frac{e^{-2}}{e^2}$$
and so
$$\lim_{x\to\infty} \left(\frac{x-2}{x+2}\right)^x = \frac{1}{e^4}.$$
A: $$\lim_{x\to\infty} (\frac{x-2}{x+2})^x$$
$$\lim_{x\to\infty} (1-\frac{4}{x+2})^x = y$$
taking log on both sides we get
$$ln(y) = x ln (1- \frac{4}{x+2})$$
the  expansion for $ln (1+r) $ is $ r- \frac{r^2}{2} +\frac{r^3}{3}$ ....
where r tends to zero
$$ln(y) = x (  \frac{-4}{x+2} - \frac{\frac{-4}{x+2}^2}{2} +\frac{\frac{-4}{x+2}^3}{3} ....)$$
$ln (y) = \frac{-4x}{x+2}$ {rest all terms will terminate to zero}
$$ln (y) =\lim_{x\to\infty} \frac{-4x}{x+2} = -4$$
$$y = \frac{1}{e^4}$$
A: you can use 
$$\left( \frac{x-2}{x+2}\right)^x = \left(1 - \frac{4}{x+2}\right)^x$$
and $(1 + \frac ax)^x \to \exp(a)$, 
HTH, AB
A: If you want to use LHopital then $ \lim_{u\to 0} \frac{\ln(1+u)}{u}=1 $ by Lhopital's rule.
 $ l= \lim_{x\to \infty} (\frac{x-2}{x+2})^x=\lim_{x\to \infty} \exp((x+2)\ln(1-\frac{4}{x+2})-2\ln(1-\frac{4}{x+2}))$
For $ u = -\frac{4}{x+2}: $ $l= \lim_{u\to 0}\exp(-4\times\frac{\ln(1+u)}{u}-2\ln(1+u))=\exp(-4)$ 
