Question

помогите пожалуйста найти предел, не используя правило Лопиталя

Answer 1

эквивалентность функции $\displaystyle \sin x\sim x,~~x\to 0$

$\displaystyle \lim_{x \to \frac{\pi}{6}}\dfrac{\sin\left(x^2-\dfrac{\pi^2}{36}\right)}{\dfrac{\sqrt{3}}{2}-\cos x}=\lim_{x \to \frac{\pi}{6}}\dfrac{x^2-\dfrac{\pi^2}{36}}{\cos\dfrac{\pi}{6}-\cos x}=\lim_{x \to \frac{\pi}{6}}\dfrac{x^2-\dfrac{\pi^2}{36}}{-2\sin\dfrac{\frac{\pi}{6}+x}{2}\sin\dfrac{\frac{\pi}{6}-x}{2}}=\\ \\ \\ =-\dfrac{1}{2}\lim_{x \to \frac{\pi}{6}}\dfrac{\left(x-\dfrac{\pi}{6}\right)\left(x+\dfrac{\pi}{6}\right)}{\sin\dfrac{\frac{\pi}{6}+x}{2}\cdot \dfrac{\frac{\pi}{6}-x}{2}}=\dfrac{1}{2}\lim_{x \to \frac{\pi}{6}}\dfrac{x+\dfrac{\pi}{6}}{\sin\dfrac{\frac{\pi}{6}+x}{2}}=\dfrac{1}{2}\cdot \dfrac{\dfrac{\pi}{6}+\dfrac{\pi}{6}}{\sin \dfrac{\pi}{6}}=\dfrac{2\pi}{3}$