Usando as propriedades de limite, calcule os seguintes limites:
\lim _{x \rightarrow \frac{\pi}{3}} \frac{\operatorname{sen}\left(x-\frac{\pi}{3}\right)}{1-2 \cos x}

Passo 1

O enunciado pede \displaystyle L=\lim_{x\to \frac{\pi}{3}}\frac{\operatorname{sen}\!\bigl(x-\tfrac{\pi}{3}\bigr)}{1-2\cos x}.
O conteúdo usado será propriedades de limites e identidades trigonométricas, em particular \cos A - \cos B = -2\operatorname{sen}\!\bigl(\tfrac{A+B}{2}\bigr)\,\operatorname{sen}\!\bigl(\tfrac{A-B}{2}\bigr)
e o limite fundamental \lim_{u\to0}\frac{\operatorname{sen} u}{u}=1.

Passo 2

Reescrevemos o denominador usando A=x e B=\tfrac{\pi}{3}:
1-2\cos x = -2\bigl(\cos x - \tfrac12\bigr) = -2\bigl(\cos x - \cos\tfrac{\pi}{3}\bigr) = -2\bigl[-2\,\operatorname{sen}\!\bigl(\tfrac{x+\pi/3}{2}\bigr)\,\operatorname{sen}\!\bigl(\tfrac{x-\pi/3}{2}\bigr)\bigr]
portanto
1-2\cos x = 4\,\operatorname{sen}\!\Bigl(\tfrac{x+\pi/3}{2}\Bigr)\,\operatorname{sen}\!\Bigl(\tfrac{x-\pi/3}{2}\Bigr).

Passo 3

O numerador também pode ser escrito em termos de ângulos metade:
\operatorname{sen}\!\bigl(x-\tfrac{\pi}{3}\bigr) =2\,\operatorname{sen}\!\Bigl(\tfrac{x-\pi/3}{2}\Bigr)\,\cos\!\Bigl(\tfrac{x-\pi/3}{2}\Bigr).
Assim a fração vira
\frac{\operatorname{sen}(x-\pi/3)}{1-2\cos x} =\frac{2\,\operatorname{sen}\!\bigl(\tfrac{x-\pi/3}{2}\bigr)\,\cos\!\bigl(\tfrac{x-\pi/3}{2}\bigr)} {4\,\operatorname{sen}\!\bigl(\tfrac{x-\pi/3}{2}\bigr)\,\operatorname{sen}\!\bigl(\tfrac{x+\pi/3}{2}\bigr)} =\frac{\cos\!\bigl(\tfrac{x-\pi/3}{2}\bigr)}{2\,\operatorname{sen}\!\bigl(\tfrac{x+\pi/3}{2}\bigr)}.

Passo 4

Agora fazemos x\to\frac{\pi}{3}. Então \tfrac{x-\pi/3}{2}\to0 e \tfrac{x+\pi/3}{2}\to\tfrac{\pi/3+\pi/3}{2}=\tfrac{\pi/3}.
Portanto,
L =\frac{\cos(0)}{2\,\operatorname{sen}(\pi/3)} =\frac{1}{2\cdot\frac{\sqrt{3}}{2}} =\frac{1}{\sqrt{3}}.

Resposta

\displaystyle \frac{1}{\sqrt{3}}

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