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? 

$$\lim_{x\to\infty} (\frac{x-2}{x+2})^x$$
$$\lim_{x\to\infty} (1-\frac{4}{x+2})^x = y$$
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}
$$ln (y) =\lim_{x\to\infty} \frac{-4x}{x+2} = -4$$
$$y = \frac{1}{e^4}$$
