Question

$\underset{x\to \infty }{lim}(\sqrt{x^2+8x+3}-\sqrt{x^2+4x+3})$ equals

• A. $0$
• B. $\infty$
• C. $2$
• D. $\frac{1}{2}$

Answer 1

The correct option is C $2$

$\underset{x\to \infty }{lim}(\sqrt{x^2+8x+3}-\sqrt{x^2+4x+3})$
$=\underset{x\to \infty }{lim}(\sqrt{x^2+8x+3}-\sqrt{x^2+4x+3}).\frac{\sqrt{x^2+8x+3}+\sqrt{x^2+4x+3}}{\sqrt{x^2+8x+3}+\sqrt{x^2+4x+3}}$
$=\underset{x\to \infty }{lim}\frac{4x}{\sqrt{x^2+8x+3}+\sqrt{x^2+4x+3}}$
$=\underset{x\to \infty }{lim}\frac{4}{\sqrt{1+8/x+3/x^2}+\sqrt{1+4/x+3/x^2}}$
$=\frac{4}{2}=2$