Question-118040

Question Number 118040 by mathocean1 last updated on 14/Oct/20

Answered by Ar Brandon last updated on 14/Oct/20

\lim_{x \to \frac{π}{3}} \frac{1 - 2 \cos x}{π - 3 x} = \frac{0}{0} = \lim_{x \to \frac{π}{3}} \frac{2 \sin x}{- 3} = 2 \times \frac{\sqrt{3}}{2} \div - 3 = - \frac{1}{\sqrt{3}}

Answered by bobhans last updated on 15/Oct/20

L = \lim_{x \to π / 3} \frac{1 - 2 \cos x}{π - 3 x}

set x = \frac{π}{3} + u, u \to 0

L = \lim_{u \to 0} \frac{1 - 2 \cos (u + \frac{π}{3})}{π - (π + 3 u)}

L = \lim_{u \to 0} \frac{1 - 2 (\frac{1}{2} \cos u - \frac{1}{2} \sqrt{3} \sin u)}{- 3 u}

L = \lim_{u \to 0} \frac{1 - \cos u + \sqrt{3} \sin u}{- 3 u}

L = \lim_{u \to 0} \frac{2 \sin^{2} (\frac{u}{2}) + 2 \sqrt{3} \sin (\frac{u}{2}) \cos (\frac{u}{2})}{- 3 u}

L = \lim_{u \to 0} \frac{2 \sin (\frac{u}{2}) {\sin (\frac{u}{2}) + \sqrt{3} \cos (\frac{u}{2})}}{- 3 u}

L = - \frac{1}{3} (0 + \sqrt{3}) = - \frac{\sqrt{3}}{3}

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