<s>[INST] 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? 
 [/INST] 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
$$
\lim_{x\rightarrow\infty}{{1\over x-2}-{1\over x+2}\over- 1/x^2}
=\lim_{x\rightarrow\infty} {-4x^2\over (x+2)(x-2)}
=\lim_{x\rightarrow\infty} {-8x\over (x+2)+(x-2)}
=\lim_{x\rightarrow\infty} {-8x\over2x}=-4.
$$
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