Question

Find the value of $\underset{x\to 0}{lim}\frac{\sqrt{a+x}-\sqrt{a}}{x}$

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

$\underset{x\to 0}{lim}\left(\frac{\sqrt{a+x}-\sqrt{a}}{x}\right)$. If we substitute x=0, we get $\frac{0}{0}$.

In the previous problems we were trying to factorize to find the limit.

When we can't factorize and there are square roots in the expression, we try to rationalize the given

expression.

$\underset{x\to 0}{lim}\frac{\sqrt{a+x}-\sqrt{a}}{x}\times \frac{\sqrt{a+x}+\sqrt{a}}{\sqrt{a+x}+\sqrt{a}}$

$=\underset{x\to 0}{lim}\frac{a+x-a}{x(\sqrt{a+x}+\sqrt{a})}$

$=\underset{x\to 0}{lim}\frac{1}{\sqrt{a+x}+\sqrt{a}}$

$=\frac{1}{\sqrt{a+0}+\sqrt{a}}$

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