You can use l'hopital's rule, since you get 0/0 by direct substitution.
Which says:
$\displaystyle \lim \frac{f(x)}{g(x)} = \lim \frac{f'(x)}{g'(x)}$
as long as the first limit is 0/0 or infinity/infinity
Therefore you get:
$\displaystyle \lim_{ x \to \1} \frac{ \sqrt{x} - 1}{x - 1} = \lim_{x\to\1} \frac{\frac{1}{2 \sqrt{x}}}{1} = \dfrac{1}{2}$
As MINOANMAN said, but let's do a quick substitution to make it look a bit nicer. Let $\displaystyle x = y^2$. Then we have
$\displaystyle \lim_{x \to 1} \frac{\sqrt{x} - 1}{x - 1} = \lim_{y \to 1} \frac{y - 1}{y^2 - 1} = \lim_{y \to 1} \frac{y - 1}{(y - 1)(y + 1)}$
etc.
-Dan