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? 

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}: $ $l= \lim_{u\to 0}\exp(-4\times\frac{\ln(1+u)}{u}-2\ln(1+u))=\exp(-4)$ 
