How can we verify this limit via $\epsilon-\delta$ method?

Solution 1:

$ \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 $\epsilon$.

Solution 2:

You can also multiply and divide by $1+\cos x$:

$$ \lim_{x\to 0}\dfrac{1-\cos x}{3 \sin^2 x} = \lim_{x\to 0}\dfrac{1-\cos^2 x}{3\sin^2 x (1+\cos x)} = \lim_{x\to 0}\dfrac{\sin^2 x}{3\sin^2 x (1+\cos x)} =\frac 13 \lim_{x\to 0}\frac{1}{1+\cos x}= \frac 16$$

If you need to use the definition, you can easily see that $$ \left| \frac{1-\cos x}{3 \sin^2 x}-\frac 16\right|=\frac 13 \left|\frac{1}{1+\cos x}-\frac 12 \right| \to 0 \quad (x \to 0) $$

Since there is no indetermination in this limit, you can use Heine's definition in a very straightforward way (instead of Cauchy's). Or, you can go on to obtain $$ \left|\frac{1}{1+\cos x}-\frac 12 \right| = \frac 12 \left|\frac{1-\cos x}{1+\cos x} \right| \leq \frac 12 |1-\cos x|\leq \frac 12 |x| $$