calculating a limit with problem with l'hopital's law I need some help in calculating this limit:
$\lim_{x\rightarrow2}(x-1)^{\frac{2x^2-8}{x^2-4x+4}}$
Thanks a lot.

HINTS:
First, note that
$$\frac{2x^2-8}{x^2-4x+4}=\frac{2(x+2)}{x-2}$$
Hence, we can write
$$(x-1)^{\frac{2(x+2)}{x-2}}=e^{\frac{2(x+2)}{x-2}\log(x-1)}$$
Finish by using L'Hospital's Rule to show 
$$\lim_{x\to 2}\frac{2(x+2)}{x-2}\log(x-1)=\lim_{x\to 2}\left(2\log(x-1)+\frac{2(x+2)}{x-1}\right)$$
