Question

lim/x→0 = arctg(−x)/x+1√−1

Answer 1

$\displaystyle \lim_{x \to 0}\dfrac{{\rm arctg}\, (-x)}{\sqrt{x+1}-1}=\{{\rm arctg}\,(-x)\sim(-x),\,\,\,x\to 0\}=\lim_{x \to 0}\dfrac{-x}{\sqrt{x+1}-1}=$

$\displaystyle-\lim_{x \to 0}\dfrac{x(\sqrt{x+1}+1)}{(\sqrt{x+1}-1)(\sqrt{x+1}+1)}=-\lim_{x \to 0}\dfrac{x\cdot(\sqrt{x+1}+1)}{x+1-1}=$

$\displaystyle-\lim_{x \to 0}\dfrac{x\cdot(\sqrt{x+1}+1)}{x}=-\lim_{x \to 0}(\sqrt{x+1}+1)=-2$