Prove some limits of cos x

Using the expansion

$\cos x = 1 - \frac{1}{2}x^2 + o(x^3) \qquad \text{as} \qquad x \to 0,$

prove that

$\frac{1-\cos x}{x^2} \to \frac{1}{2} \qquad \text{as} \qquad x \to 0.$

Using a similar method, find

$\lim_{x \to 0} \frac{1-\cos (2x) - 2x^2}{x^4}.$

Proof. Since

$\cos x = 1 - \frac{1}{2}x^2 + o(x^3)$

we have

$\begin{align*} &&1 - \cos x &= \frac{1}{2} x^2 + o(x^3) \\[9pt] \implies && \frac{1 - \cos x}{x^2} &= \frac{1}{2} + o(x) &(\text{Theorem 7.8(c)}) \\[9pt] \implies && \frac{1 - \cos x}{x^2} &\to \frac{1}{2} \quad \text{as} \quad x \to 0. \end{align*}$

(We use Theorem 7.8(c) since in the second line to bring the $\frac{1}{x^2}$ inside the little $o(x^3)$ .)

For the second limit, we use the Taylor expansion for $\cos x$ and replace $x$ with $2x$ to obtain,

$\begin{align*} &&\cos (2x) &= 1 - 2x^2 + \frac{2}{3}x^4 + o(x^5) \\[9pt] \implies && 1 - \cos (2x) - 2x^2 &= 1 - 1 + 2x^2 - \frac{2}{3}x^4 + 2x^2 + o(x^5) \\[9pt] \implies && 1 - \cos (2x) - 2x^2 &= -\frac{2}{3}x^4 + o(x^5) \\[9pt] \implies && \frac{1 - \cos (2x) - 2x^2}{x^4} &\to -\frac{2}{3} \quad \text{as} \quad x \to 0. \qquad \blacksquare \end{align*}$

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