Question

${lim}_{x\to \infty }(\sqrt{x^2-x+1}-ax-b)=0,$ then the values of $a$ and $b$ are given by

• A. $a=-1,b=1/2$
• B. $a=1,b=1/2$
• C. $a=1,b=-1/2$
• D. None of these

Answer 1

The correct option is C $a=1,b=-1/2$

We have,

$\begin{array}{c}{lim}_{x\to \infty }\left(\sqrt{x^2-x+1}-ax-b\right)=0\\ Putx=\frac{1}{t}\\ {lim}_{t\to 0^{+}}\left(\sqrt{\frac{1}{t^2}-\frac{1}{t}+1}-\frac{a}{t}-6\right)=0\\ {lim}_{x\to 0^{+}}\frac{{\left(1-t+t^2\right)}^{\frac{1}{2}}-a-bt}{t}=0\\ {lim}_{x\to 0^{+}}\frac{\left(1-\frac{t}{2}+\frac{t^2}{2}\right)-a-bt}{t}=0\\ {lim}_{x\to 0^{+}}\frac{\left(1-a\right)+t\left(-b-\frac{1}{2}\right)}{t}=0\\ 1=a\\ andb=-\frac{1}{2}\\ OptionCiscorrectanswer.\end{array}$