Question

Solve $\underset{x\to \infty }{\lim }\left(\sqrt{x^2+ax+b}-x\right)=$

• A. $\frac{b}{2}$
• B. $\frac{a}{2}$
• C. $-\frac{b}{2}$
• D. $-\frac{a}{2}$

Answer 1

Answer: (B)

$\frac{a}{2}$

$\underset{x\to \infty }{lim}\sqrt{x^2+ax+b}-x$

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

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

$=\underset{x\to \infty }{lim}\frac{ax+b}{\sqrt{x^2+ax+b+x}}$

$=\underset{x\to \infty }{lim}\frac{a+b/x}{\sqrt{1+\frac{a}{x}+\frac{b}{x^2}}+1}$

$=\frac{a}{2}$