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.

It's probably better to do a change of variable, $x=t+2$, so the exponent becomes
$$
\frac{2x^2-8}{x^2-4x+4}=\frac{2(x-2)(x+2)}{(x-2)^2}=2\frac{x+2}{x-2}=2\frac{t+4}{t}
$$
Thus the limit becomes
$$
\lim_{t\to0}\bigl((1+t)^{1/t}\bigr)^{2(t+4)}
$$
Now it's a matter of computing
$$
l=\lim_{t\to0}(1+t)^{1/t}
$$
and then the sought limit is $l^8$.
