Say I have to calculate the limit
$$\lim_{x\to 0} \dfrac{1 - \cos(x)}{3\sin^2(x)}$$
Now, from the use of De l'Hôpital rule (or other methods like notable special limits), we know the limit is $\dfrac{1}{6}$.
Let's say we want to verify it by the definition. Hence
$$\big|\dfrac{1 - \cos(x)}{3\sin^2(x)} - \frac{1}{6}\big| < \epsilon$$
$$\big|\dfrac{2 - 2\cos(x) - \sin^2(x)}{6\sin^2(x)}\big| < \epsilon$$
What would come into my mind is to use estimations but this wouldn't work because I am not treating a succession (where $n\to+\infty$ hence I can use $|\sin(n)| = 1$ and similar trick).
So how would one reason with a case like this?
I tried to rewrite the numerator in cosine terms only, and then I managed to rewrite it as
$$1 - 2\cos(x) + \cos^2(x) \to (\cos(x) - 1)^2$$
forgetting about periodicity of the solutions (the associate equation $y^2 - 2y + 1 = 0$ would give $t = 1$ hence $x = \frac{\pi}{4} + 2k\pi$ as a solution), so:
$$\big|\dfrac{(\cos(x)-1)^2}{6\sin^2(x)}\big| < \epsilon$$
But now the absolute value is meaningless since numerator and denominator are always positive (?), so how to proceed?
$ \newcommand{\R}{\mathbb{R}} $ We have to show that $$ \forall \epsilon > 0: \exists \delta \in (0, \pi/2]: \forall x \in \R\setminus\{0\}: (|x| < \delta \Rightarrow |f(x)| < \epsilon) $$ where $$ f(x) = \dfrac{(1 - \cos x)^2}{6 \sin^2 x} = \dfrac{(1 - \cos x)^2}{6(1- \cos^2 x)} $$ Let $ t = \cos x$. Then, because $ t \neq 1 $ whenever $x \neq 0$ and $ |x| < \pi/2 $,
$$ f(x) = \dfrac{(1 - t)^2}{6(1- t^2)} = \dfrac{(1 - t)}{6(1 + t)} = \dfrac{(2 -1 - t)}{6(1 + t)} = \dfrac{1}{3(1 + t)} - \dfrac{1}{6} $$ $f(x)$ strictly decreases w.r.t. $t$.
Find $t$ that makes $f(x) = \epsilon$. After all, $t = 2/(6\epsilon + 1) - 1$. Since $ t = \cos x$, choosing
$$ \delta = \arccos \left(\dfrac{2}{6\min\{\epsilon, 1/6\} + 1} - 1 \right) $$ proves the claim. I think you can show that $\delta$ is an increasing function of $\epsilon$.