While deriving $\frac{\rm{d}}{\rm{dx}}\rm{sin}(x)$, using the definition of the derivative and expanding $\rm{sin}(x+h)$ leads to

$\frac{\rm{d}}{\rm{dx}}\rm{sin}(x) = \rm{sin}(x)\lim\_{h\to 0} \frac{cos(h)-1}{h} + \rm{cos}(x)lim\_{h\to 0}\frac{\rm{sin}(h)}{h}$

The second limit can be evaluated by applying the squeeze theorem, and if I wasn't going from first principles I could use l'Hopital's rule for the first limit, but that would be circular logic.

How can I evaluate $\lim\_{h\to 0} \frac{cos(h)-1}{h}$ without l'Hopital?