Question

${\lim }_{x\to \infty }\left[\sqrt{x^2+ax+b}-x\right]\left(a<0<b\right)$

• A. depends on both $a$ and $b$
• B. depends only on $b$
• C. depends only on $a$
• D. does not depend on $a$ and $b$

Answer 1

Answer: (C)

Given that, $\underset{x\to \infty }{\lim }\left[\sqrt{x^2+ax+b}-x\right]$
$=\underset{x\to \infty }{\lim }\left(\sqrt{x^2+ax+b}-x\right)\times \left(\frac{\sqrt{x^2+ax+b}+x}{\sqrt{x^2+ax+b}+x}\right)$
[By rationalisation]
$=\underset{x\to \infty }{\lim }\frac{x^2+ax+b-x^2}{\sqrt{x^2+ax+b}+x}=\underset{x\to \infty }{\lim }\frac{ax+b}{\sqrt{x^2+ax+b}+x}$
$=\underset{x\to \infty }{\lim }\frac{x\left(a+\frac{b}{x}\right)}{x(\sqrt{1+\frac{a}{x}+\frac{b}{x^2}+1)}}$
(taking $x$ common from both numerator and denominator)
$=\frac{a}{(1+1)}=\frac{a}{2}$
It is clear that it depends only on $a$