Find out the answer to "compute the limit oflimx→π31−2cos⁡(x)sin⁡(3x)I would like to not..."

Coradossi7xod

Answered question

2022-03-27

compute the limit of

lim_{x \to \frac{π}{3}} \frac{1 - 2 \cos (x)}{\sin (3 x)}

I would like to not do a translation with the change of variable

t = x - \frac{π}{3}

diocedss33

Beginner2022-03-28Added 12 answers

$\frac{1 - 2 \cos x}{\sin (3 x)} \frac{\sqrt{3}}{3}$

$= \frac{1 - 2 \cos (x)}{- 4 \sin^{3} x + 3 \sin x}$

$= \frac{1}{\sin x} \frac{1 - 2 \cos (x)}{3 - 4 \sin^{2} x}$

$= \frac{1}{\sin x} \frac{1 - 2 \cos (x)}{3 - 4 + 4 \cos^{2} x}$

$= \frac{1}{\sin x} \frac{1 - 2 \cos (x)}{4 \cos^{2} x - 1}$

$= \frac{1}{\sin x} \frac{1 - 2 \cos (x)}{(2 \cos x - 1) (2 \cos x + 1)}$

$= \frac{1}{\sin x} \frac{- 1}{(2 \cos x + 1)}$

$\to \frac{2}{\sqrt{3}} \frac{- 1}{2}$

$= \frac{- 1}{\sqrt{3}}$

$= - \frac{\sqrt{3}}{3}$

Roy Brady

Beginner2022-03-29Added 19 answers

Since we have a case of

\frac{0}{0}

we use L'Hospital rule.

lim_{x \to \frac{π}{3}} \frac{1 - 2 \cos (x)}{\sin (3 x)} = lim_{x \to \frac{π}{3}} \frac{2 \sin (x)}{3 \cos (3 x)} = - \frac{\sqrt{3}}{3}

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