In the previous problem I proved that:
$$\lim_{x\to 0}\frac{\sin(x)}{x}=1$$
Using the following inequality: $$\cos x \le\frac{x}{\sin x}\le\frac{1}{\cos x}$$
which was derived geometrically using the area of circles and triangles. The problem I am stuck on is proving the following:
$$\lim_{x\to 0}\frac{\cos x-1}{x}=0$$
using the first formula. I have tried just rewriting the first expression to this one using the trig identities but have not been able to succeed yet. There is no solution in my textbook. Could someone guide me in the right direction without giving a full solution?
