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? 

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
