Question

The value of $\underset{n\to \infty }{lim}\frac{\sqrt{3n^2-1}-\sqrt{2n^2-1}}{4n+3}$ is

• A. $\frac{1}{4}(\sqrt{3}-\sqrt{2})$
• B. $\frac{1}{4}(\sqrt{3}+\sqrt{2})$
• C. $(\sqrt{3}-\sqrt{2})$
• D. none of these

Answer 1

The correct option is A $\frac{1}{4}(\sqrt{3}-\sqrt{2})$

$\underset{n\to \infty }{lim}\frac{\sqrt{3n^2-1}-\sqrt{2n^2-1}}{4n+3}$

$=\underset{n\to \infty }{lim}\frac{\sqrt{3n^2-1}-\sqrt{2n^2-1}}{4n+3}.\frac{\sqrt{3n^2-1}+\sqrt{2n^2-1}}{\sqrt{3n^2-1}+\sqrt{2n^2-1}}$

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

$=\underset{n\to \infty }{lim}\frac{n}{4n+3}.\frac{n}{\sqrt{3n^2-1}+\sqrt{2n^2-1}}$

$=\underset{n\to \infty }{lim}\frac{1}{4+\frac{3}{n}}.\frac{1}{\sqrt{3-\frac{1}{n^2}}+\sqrt{2-\frac{1}{n^2}}}$

$=\frac{1}{4}.\frac{1}{\sqrt{3}+\sqrt{2}}=\frac{1}{4}(\sqrt{3}-\sqrt{2})$