Question

$\underset{x\to 3}{lim}=\frac{\sqrt{x}-\sqrt{3}}{\sqrt{x^2-9}}$ is equal to

• A. $1$
• B. $3$
• C. $\sqrt{3}$
• D. $-\sqrt{3}$
• E. $0$

Answer 1

Answer: (E)

$0$

The value of $\underset{x\to 3}{lim}=\frac{\sqrt{x}-\sqrt{3}}{\sqrt{x^2-9}}$ is

$=\underset{x\to 3}{lim}\frac{\sqrt{x}-\sqrt{3}}{\sqrt{x-3}\sqrt{x}+3}\times \frac{\sqrt{x}+\sqrt{3}}{\sqrt{x}+\sqrt{3}}$
$=\underset{x\to 3}{lim}\frac{x-3}{\sqrt{x-3}\sqrt{x+3}}\times \frac{1}{\sqrt{x}+\sqrt{3}}$
$=\underset{x\to 3}{lim}\frac{\sqrt{x-3}}{\sqrt{x+3}}\times \frac{1}{\sqrt{x}+\sqrt{3}}=0$