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[tex]\lim_{x \to \ \frac{\pi }{6} } \frac{sin(x-\frac{\pi }{6}) }{\frac{\sqrt{3} }{2} - cosx}[/tex]

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NNNLLL54

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Ответ:

$\lim\limits_{x \to \frac{\pi }{6}}\dfrac{sin\Big(x-\dfrac{\pi }{6}\Big)}{\dfrac{\sqrt3}{2}-cosx}=\Big[\ t=x-\dfrac{\pi }{6}\ ,\ x=t+\dfrac{\pi }{6}\ \Big]=\lim\limits_{t \to 0}\dfrac{sint}{\dfrac{\sqrt3}{2}-cos\Big(t+\dfrac{\pi }{6}\Big)}=\\\\\\=\lim\limits_{t \to 0}\dfrac{sint}{\dfrac{\sqrt3}{2}-cos\, t\cdot cos\dfrac{\pi }{6}+sint\cdot sin\dfrac{\pi }{6}}=\lim\limits_{t \to 0}\dfrac{sint}{\dfrac{\sqrt3}{2}-cost\cdot \dfrac{\sqrt3}{2}+sint\cdot \dfrac{1}{2}}=$

$=\lim\limits_{t \to 0}\dfrac{2sin\dfrac{t}{2}\cdot cos\dfrac{t}{2}}{\dfrac{\sqrt3}{2}\Big(1-cost\Big)+sin\dfrac{t}{2}\cdot cos\dfrac{t}{2}}=\lim\limits_{t \to 0}\dfrac{2sin\dfrac{t}{2}\cdot cos\dfrac{t}{2}}{\dfrac{\sqrt3}{2}\cdot 2sin^2\dfrac{t}{2}+sin\dfrac{t}{2}\cdot cos\dfrac{t}{2}}=$

$=\lim\limits_{t \to 0}\dfrac{2\cdot cos\dfrac{t}{2}}{\sqrt3\cdot sin\dfrac{t}{2}+cos\dfrac{t}{2}}=\dfrac{2\cdot 1}{\sqrt3\cdot 0+1}=2$

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