[tex]\frac{1}{2}=lim(x\rightarrow -\infty)\frac{a\sqrt{x^2+1}+2017}{x+2018}=lim(x\rightarrow -\infty)\frac{-ax\sqrt{1+\frac{1}{x^2}}+2017}{x+2018}=lim(x\rightarrow -\infty)\frac{-a\sqrt{1+\frac{1}{x^2}}+\frac{2017}{x}}{1+\frac{2018}{x}}=-a => 4a=-2[/tex]
[tex]1=lim(\sqrt{x^2+bx+1}-x)=lim\frac{bx+1}{\sqrt{x^2+bx+1}+x}=lim\frac{b+\frac{1}{x}}{\sqrt{1+\frac{b}{x}+\frac{1}{x^2}}+1}=\frac{b}{2}=> b=2[/tex]