Prove $\lim_{x \to a} \cos{x} = \cos{a}$ using a $\varepsilon$-$\delta$ argument

Question

Prove $\lim\limits_{x \to a} \cos{x} = \cos{a}$ with $\varepsilon$-$\delta$.

Is proving $|\cos{x}-\cos{a}| \leq |x-a|$ with MVT the only way possible?

Answer 1

Hint: Use $$\cos(x) - \cos(a) = -2\sin\left(\frac{x+a}{2}\right)\sin\left(\frac{x-a}{2}\right)$$

Answer 2

Let $\varepsilon > 0$ be given and suppose $\delta = \varepsilon$. Let $x$ satisfy $0<|x−a| < \delta$, then

\begin{align}|\cos x-\cos a|&=\left|-2\sin\left(\frac{x+a}{2}\right)\sin\left(\frac{x-a}{2}\right) \right|\\&= 2\left|\sin\left(\frac{x+a}{2}\right)\right|\cdot\left|\sin\left(\frac{x-a}{2}\right)\right|\\&\le 2\left|\frac{x-a}{2}\right|. \end{align}

Now conclude.