Question

The values of constants $a$ and $b$ so that $\underset{x\to \infty }{lim}(\frac{x^2+1}{x+1}-ax-b)=0$

Answer 1

$\begin{array}{c}\Rightarrow {lim}_{x\to \propto }\left(\frac{x^2+1-a{x}^2-bx-ax-b}{x+1}\right)=0\\ \Rightarrow {lim}_{x\to \propto }{x}^2\frac{\left\{\left(1-a\right)-\frac{\left(a+b\right)}{x}-\frac{b}{x^2}\right\}}{x^2\left(x+\frac{1}{x^2}\right)}=0\\ \Rightarrow Itis\frac{0}{0}form,1-a=0,a=1\\ a+b=0,b=-1Ans.\end{array}$