Osvaldo Apodaca

2021-12-30

How can I evaluate $\underset{v\to \frac{\pi }{3}}{lim}\frac{1-2\mathrm{cos}v}{\mathrm{sin}\left(v-\frac{\pi }{3}\right)}$ without using LHospitals rule?

Medicim6

You can use the fact that
$2\mathrm{cos}v=2\mathrm{cos}\left(\left(v-\frac{\pi }{3}\right)+\frac{\pi }{3}\right)$
$=\mathrm{cos}\left(v-\frac{\pi }{3}\right)-\sqrt{3}\mathrm{sin}\left(v-\frac{\pi }{3}\right)$
It follows from this that
$\frac{1-2\mathrm{cos}v}{\mathrm{sin}\left(v-\frac{\pi }{3}\right)}=\frac{1-\mathrm{cos}\left(v-\frac{\pi }{3}\right)}{\mathrm{sin}\left(v-\frac{\pi }{3}\right)}-\sqrt{3}$
and therefore all that remains to be done is to compute
$\underset{t\to 0}{lim}\frac{1-\mathrm{cos}t}{\mathrm{sin}t}$
and for this you can use the fact that

temnimam2

just use Taylor expansions:

$\underset{x\to \frac{\pi }{3}}{lim}\frac{1-2\mathrm{cos}x}{\mathrm{sin}\left(x-\frac{\pi }{3}\right)}=\underset{x\to \frac{\pi }{3}}{lim}\frac{1-1+\sqrt{3}\left(x-\frac{\pi }{3}\right)+o\left(x-\frac{\pi }{3}\right)}{x-\frac{\pi }{3}+o\left(x-\frac{\pi }{3}\right)}=\underset{x\to \frac{\pi }{3}}{lim}\sqrt{3}+o\left(1\right)=\sqrt{3}$

Vasquez

Use Prosthaphaeresis fromula and $\mathrm{sin}2A=2\mathrm{sin}A\mathrm{cos}A$ Formulas,
$2\cdot \underset{v\to \frac{\pi }{3}}{lim}\frac{\mathrm{cos}\frac{\pi }{3}-\mathrm{cos}v}{\mathrm{sin}\left(v-\frac{\pi }{3}\right)}=2\cdot \underset{v\to \frac{\pi }{3}}{lim}\frac{2\mathrm{sin}\left(\frac{v}{2}-\frac{\pi }{6}\right)\mathrm{sin}\left(\frac{v}{2}+\frac{\pi }{6}\right)}{2\mathrm{sin}\left(\frac{v}{2}-\frac{\pi }{6}\right)\mathrm{cos}\left(\frac{v}{2}-\frac{\pi }{6}\right)}$
Now cancel out

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