Question

Find the constants a and b such that $\underset{x\to \infty }{lim}(\sqrt{x^2-x+1}-ax-b)=0$

Answer 1

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

let $\frac{1}{x}=t$

$x\to \infty \Rightarrow t\to 0$

$\underset{t\to 0}{lim}(\sqrt{\frac{1}{t^2}-\frac{1}{t}+1}-\frac{a}{t}-b)=0$

$\underset{t\to 0}{lim}(\frac{\sqrt{t^2-t+1}}{t}-\frac{a}{t}-b)=0$

$\Rightarrow \underset{t\to 0}{lim}\frac{\sqrt{1+(t^2-t)}-a-bt}{t}=0$

$\Rightarrow \underset{t\to 0}{lim}\frac{1+(t^2-t)-a-bt}{t}=0[using{(1+x)}^n=1+nxforx\to 0]$

$\Rightarrow \underset{t\to 0}{lim}\frac{(1-a)+(-1-b)t+t^2}{t}=0$

For time to exist

$1-a=0\Rightarrow a=1$

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

$=-(1+b)=0$ (given)

$b=-1$