Question

Решите предел не пользуясь правилом Лопиталя

Answer 1

$\displaystyle \lim_{x \to 1}\dfrac{\sqrt{x^2-3x+3}-1}{\sin\pi x}=\lim_{x \to 1}\dfrac{\Big(\sqrt{x^2-3x+3}-1\Big)\Big(\sqrt{x^2-3x+3}+1\Big)\pi x}{\pi x\sin\pi x\Big(\sqrt{x^2-3x+3}+1\Big)}=\\ \\ \lim_{x \to 1}\dfrac{x^2-3x+3-1}{\pi \cdot \Big(\sqrt{1^2-3\cdot 1+3}+1\Big)}=\dfrac{1^2-3\cdot 1+2}{2\pi}=0$