Why is $1 - \cos(x)$ of $\mathcal{O}(x^2)$?

Question

I know the definition of order estimates,

For the solution to apply we need to show:

$ \displaystyle\lim_{x \to 0} \frac{1 - \cos(x)}{x^2} = A \neq 0\space or \space \infty$

But how can one show this?

Answer 1

Hint: If you already know that $\displaystyle\lim_{x \to 0}\dfrac{\sin x}{x} = 1$, then you can use the following: $\dfrac{1-\cos x}{x^2} = \dfrac{1-\cos x}{x^2} \cdot \dfrac{1+\cos x}{1+\cos x} = \dfrac{1-\cos^2 x}{x^2(1+\cos x)} = \dfrac{\sin^2 x}{x^2(1+\cos x)} = \left[\dfrac{\sin x}{x}\right]^2\dfrac{1}{1+\cos x}$.