zagonek34

2022-01-12

How can we verify this limit via $ϵ-\delta$ method?
I have to calculate the limit
$\underset{x\to 0}{lim}\frac{1-\mathrm{cos}\left(x\right)}{3{\mathrm{sin}}^{2}\left(x\right)}$

Lakisha Archer

We have to show that

$:\left(|x|<\delta ⇒|f\left(x\right)|<ϵ\right)$
where $f\left(x\right)=\frac{{\left(1-\mathrm{cos}x\right)}^{2}}{6{\mathrm{sin}}^{2}x}=\frac{{\left(1-\mathrm{cos}x\right)}^{2}}{6\left(1-{\mathrm{cos}}^{2}x\right)}$
Let $t=\mathrm{cos}x$. Then, because ,
$f\left(x\right)=\frac{{\left(1-t\right)}^{2}}{6\left(1-{t}^{2}\right)}=\frac{\left(1-t\right)}{6\left(1+t\right)}=\frac{\left(2-1-t\right)}{6\left(1+t\right)}$
$=\frac{1}{3\left(1+t\right)}-\frac{1}{6}$
f(x) strictly decreases w.r.t.t.
Find t that makes $f\left(x\right)=ϵ$. After all, $t=\frac{2}{\left(6ϵ+1\right)}-1$. Since $t=\mathrm{cos}x$, choosing
$\delta =\mathrm{arccos}\left(\frac{2}{6min\left\{ϵ,\frac{1}{6}\right\}+1}-1\right)$
proves the claim. I think you can show that $\delta$ is an increasing function of $ϵ$.

Mary Nicholson

You can also multiply and divide by $1+\mathrm{cos}x$:
$\underset{x\to 0}{lim}\frac{1-\mathrm{cos}x}{3{\mathrm{sin}}^{2}x}=\underset{x\to 0}{lim}\frac{1-{\mathrm{cos}}^{2}x}{3{\mathrm{sin}}^{2}x\left(1+\mathrm{cos}x\right)}=$
$\underset{x\to 0}{lim}\frac{{\mathrm{sin}}^{2}x}{3{\mathrm{sin}}^{2}x\left(1+\mathrm{cos}x\right)}=\frac{1}{3}\underset{x\to 0}{lim}\frac{1}{1+\mathrm{cos}x}=\frac{1}{6}$
If you need to use the definition, you can easily see that
$|\frac{1-\mathrm{cos}x}{3{\mathrm{sin}}^{2}x}-\frac{1}{6}|=\frac{1}{3}|\frac{1}{1+\mathrm{cos}x}-\frac{1}{2}|\to 0\left(x\to 0\right)$
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
$|\frac{1}{1+\mathrm{cos}x}-\frac{1}{2}|=\frac{1}{2}|\frac{1-\mathrm{cos}x}{1+\mathrm{cos}x}|\le \frac{1}{2}|1-\mathrm{cos}x|\le \frac{1}{2}|x|$

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