Question

с пределами. без Лопиталя! ​

Answer 1

$\lim\limits_{x \to \frac{\pi }{3}}\dfrac{1-2cosx}{\pi -3x}=\Big[\ t=x-\frac{\pi}{3}\ ;\ x=t+\dfrac{\pi}{3}\Big]=\lim\limits_{t \to 0}\dfrac{1-2cos(t+\dfrac{\pi}{3})}{-3t}=\\\\\\=\lim\limits_{t \to 0}\dfrac{1-2\cdot (cost\cdot cos\dfrac{\pi}{3}-sint\cdot sin\dfrac{\pi}{3})}{-3t}=\lim\limits_{t \to 0}\dfrac{1-cost+\sqrt3\, sint}{-3t}=$

$=\lim\limits_{t \to 0}\dfrac{1-cost}{-3\, t}+\lim\limits_{t \to 0}\dfrac{\sqrt3\, sint}{-3t}=\lim\limits_{t \to 0}\dfrac{2\, sin^2\dfrac{t}{2}}{-3t}-\lim\limits_{t \to 0}\dfrac{\sqrt3\, sint}{3t}=\\\\\\=\Big[\ sin\alpha (x)\sim \alpha (x)\ ,\ \alpha (x)\to 0\ \Big]=\lim\limits_{t \to 0}\dfrac{2\cdot \Big(\dfrac{t}{2}\Big)^2}{-3\cdot t}-\lim\limits_{t \to 0}\dfrac{\sqrt3\cdot t}{3\cdot t}=$

$=\lim\limits_{t \to 0}\, \dfrac{2\cdot t^2}{-3\cdot t\cdot 4}-\lim\limits_{t \to 0}\dfrac{\sqrt3\cdot t}{3\cdot t}=\lim\limits_{t \to 0}\, \dfrac{t}{-6}-\lim\limits_{t \to 0}\dfrac{\sqrt3}{3}=0-\dfrac{\sqrt3}{3}=-\dfrac{\sqrt3}{3}$