Question

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

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

Answer 1

The correct option is C $1$

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

Rationalizing

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

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

$=\underset{x\to \infty }{lim}\frac{2}{(\sqrt{1+1/x}+\sqrt{1-1/x})}=\frac{2}{1+1}=1$

$(asx\to \infty ;\frac{1}{x}\to 0)$