Question

Решить, заранее спасибо :)

Answer 1

$lim_{x\to0}\frac{\sqrt{1-x^2}-1}{x}=lim_{x\to0}\frac{(\sqrt{1-x^2}-1)(\sqrt{1-x^2}+1)}{x(\sqrt{1-x^2}+1)}=lim_{x\to0}\frac{1-x^2-1}{x(\sqrt{1-x^2}+1)}=\\=lim_{x\to0}\frac{-x^2}{x(\sqrt{1-x^2}+1)}=lim_{x\to0}\frac{-x}{\sqrt{1-x^2}+1}=\frac{0}{\sqrt{1-0}+1}=0$


$lim_{x\to2-0}\frac{tg(2-x)}{2-x}=lim_{x\to2-0}\frac{tg(2-x)}{2-x}=lim_{x\to2-0}\frac{sin(2-x)}{(2-x)cos(2-x)}=\\=lim_{x\to2-0}\frac{sin(2-x)}{2-x}*lim_{x\to2-0}\frac{1}{cos(2-x)}=1*\frac{1}{cos(2-2)}=1*\frac{1}{1}=1$