Question

Find the value of $a$ + $b$ if $\underset{x\to \infty }{lim}(\sqrt{x^2-x+1}-ax-b)=0,a>0.$

Answer 1

$\underset{x\to \infty }{lim}(\sqrt{x^2-x+1}-ax-b)=0$

$\underset{x\to \infty }{lim}(x\sqrt{\frac{x^2}{x^2}-\frac{x}{x^2}+\frac{1}{x^2}}-\frac{ax}{x}-\frac{b}{x})=0$

$\underset{x\to \infty }{lim}[\sqrt{1-\frac{1}{x}+\frac{1}{x^2}}-a-\frac{b}{x}]=0$

so $\sqrt{1}-a-0=0$ (we know $\underset{x\to \infty }{lim}=\frac{1}{x}=0$)

$a=1$

put this value

$\sqrt{x^2-x+1}-ax-b=0$

$\sqrt{1-1+1}-a-b=0$

$1-1-b=0$

$b=0$

so $a=1$ & $b=0$.