Question

The value of $\underset{x\to \infty }{lim}(\sqrt{3x^2+\sqrt{3x^2+\sqrt{3x^2}}}-\sqrt{3x^2})$ is

• A. $\frac{1}{2}$
• B. $-\frac{1}{2}$
• C. $-\frac{3}{2}$
• D. $\sqrt{3}$

Answer 1

The correct option is A $\frac{1}{2}$

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

Rationalize

$\underset{x\to \infty }{lim}\frac{{\left(\sqrt{3x^2+\sqrt{3x^2+\sqrt{3x^2}}}\right)}^2-(\sqrt{3x}{)}^2}{\sqrt{3x^2+\sqrt{3x^2+\sqrt{3x^2}}}+\sqrt{3x^2}}$

$=\underset{x\to \infty }{lim}\frac{\sqrt{3x^2+\sqrt{3x^2}}}{\sqrt{3x^2+\sqrt{3x^2+\sqrt{3x^2}}}+\sqrt{3x^2}}$

$=\underset{x\to \infty }{lim}\frac{|x|\sqrt{3+\sqrt{3/x^2}}}{|x|\sqrt{3+\frac{\sqrt{3x^2+\sqrt{3x^2}}}{x^2}+\sqrt{3}|x|}}$

$=\frac{\sqrt{3+0}}{\sqrt{3+0}+\sqrt{3}}$

$=\frac{\sqrt{3}}{2\sqrt{3}}=\frac{1}{2}$

Option $\left(A\right)$