<s>[INST] To find the value of a function at a point where it is continuous Find $f(0)$ so that the function $f(x)=\dfrac{1-\cos(1-\cos x)}{x^4}$ is continuous everywhere.
My attempt:
By applying sandwich theorem $-1 \le \cos(x) \le 1$.
$$1 \ge -\cos(x) \ge -1$$
$$2 \ge 1-\cos(x) \ge 0$$
$\cos(x)$ is a decreasing function from $0^\circ$ to $90^\circ$.
$$\cos(2) \le \cos(1-\cos(x)) \le 1$$
$$-\cos(2) \ge -\cos(1-\cos(x)) \ge -1$$
$$1-\cos(2) \ge 1-\cos(1-\cos(x)) \ge 0$$
$$\frac{(1-\cos(2))}{(x^4)} \ge \frac{(1-\cos(1-\cos(x)))}{(x^4)} \ge 0$$
Thus I got the answer 0 applying limits every where.
But the actual answer is 1/8 apparently. How?
 [/INST] It is enough to note that $\lim_{y \to 0} \frac{1-\cos y}{y^2}=\frac{1}{2}$.
Now you write
$$ \frac{1-\cos(1-\cos x)}{x^4}= \frac{1-\cos(1-\cos x)}{(1-\cos x)^2}\left( \frac{1-\cos x}{x^2}\right)^2 \to \frac{1}{2} \frac{1}{4}=\frac{1}{8}$$
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