How can I evaluate limv→π31−2cos⁡vsin⁡(v−π3) without using LHospitals rule?

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Medicim6 · Accepted Answer

You can use the fact that  2cos⁡v=2cos⁡((v−π3)+π3)  =cos⁡(v−π3)−3sin⁡(v−π3)  It follows from this that  1−2cos⁡vsin⁡(v−π3)=1−cos⁡(v−π3)sin⁡(v−π3)−3  and therefore all that remains to be done is to compute  limt→01−cos⁡tsin⁡t  and for this you can use the fact that  1−cos⁡t=2sin2(t2)   and that  sin⁡t=2sin⁡(t2)cos⁡(t2)

temnimam2 · Accepted Answer

just use Taylor expansions:  cos⁡x=12−3⋅(x−π3)2+o(x−π3), x→π3  sin⁡x=x−π3, o(x−π3), x→π3  limx→π31−2cos⁡xsin⁡(x−π3)=limx→π31−1+3(x−π3)+o(x−π3)x−π3+o(x−π3)=limx→π33+o(1)=3

Vasquez · Accepted Answer

Use Prosthaphaeresis fromula and sin⁡2A=2sin⁡Acos⁡A Formulas, 2⋅limv→π3cos⁡π3−cos⁡vsin⁡(v−π3)=2⋅limv→π32sin⁡(v2−π6)sin⁡(v2+π6)2sin⁡(v2−π6)cos⁡(v2−π6) Now cancel out sin⁡(v2−π6) as sin⁡(v2−π6)≠0 as v→π3, v≠π3

How can I evaluate \lim_{v \to \frac{\pi}{3}} \frac{1-2\cos v}{\sin(v-\frac{\pi}{3})} without

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