Question

The value of the limit $\underset{x\to \infty }{\lim }\left(\sqrt{x+3}-\sqrt{x}\right\}\sqrt{x+1}$ is

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

Answer 1

The correct option is C $\frac{3}{2}$

$\underset{x\to \infty }{\lim }\left(\sqrt{x+3}-\sqrt{x}\right\}\sqrt{x+1}=\underset{x\to \infty }{\lim }\frac{\left(\sqrt{x+3}-\sqrt{x}\right\}\left(\sqrt{x+3}+\sqrt{x}\right\}\sqrt{x+1}}{\left(}=\underset{x\to \infty }{\lim }\frac{\left(x+3-x\right)\frac{\sqrt{x+1}}{\sqrt{x}}}{\frac{\sqrt{x+3}+\sqrt{x}}{\sqrt{x}}}=3\underset{x\to \infty }{\lim }\frac{\sqrt{1+\frac{1}{x}}}{\sqrt{1+\frac{3}{x}}+1}\mathrm{As}x\to \infty ,\frac{1}{x}\to 0=3\times \frac{1}{2}=\frac{3}{2}$