Question

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

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

Answer 1

The correct option is B $\frac{a}{2}$

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

Multypliying and divinding $(\sqrt{x^2+ax+b}+x)$ we get,

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

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

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

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

$\Rightarrow \frac{a+\frac{b}{\infty }}{\sqrt{1+\frac{a}{\infty }+\frac{b}{{\infty }^2}}+1}$

$\Rightarrow \frac{a}{2}$