Prove $\lim \frac{cos(x)-1}{x} = 0$ without l'Hopital

Question

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?

Answer 1

$$\frac{\cos(h)-1}{h}=\frac{\cos(h)-\cos(0)}{h-0} \longrightarrow \cos'(0)=0$$

(if you don't know yet that $\cos'=-\sin$, you can still get that $\cos'(0)=0$ by noticing that $\cos$ has a local maximum at $x=0$, since $\cos(0)=1=\max_{x \in \mathbb{R}} \cos(x)$)

Answer 2

You can do $$ \frac{\cos h-1}h=\frac{\cos^2h-1}{h(\cos h+1)}=-\sin h \,\frac{\sin h}h\,\frac1{\cos h+1}. $$ The nontrivial one is always $\displaystyle\frac{\sin h}h$.