Question

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

Answer 1

$\underset{x\to 0}{\lim }\left[\frac{\sqrt{1+3x}-\sqrt{1-3x}}{x}\right]$

It is of the form $\frac{0}{0}$.

Rationalising the numerator:

$$\begin{gathered} \underset{x\to 0}{\lim }\left[\frac{\left(\sqrt{1+3x}-\sqrt{1-3x}\right)\left(\sqrt{1+3x}+\sqrt{1-3x}\right)}{x\left(\sqrt{1+3x}+\sqrt{1-3x}\right)}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{\left(1+3x\right)-\left(1-3x\right)}{x\left(\sqrt{1+3x}+\sqrt{1-3x}\right)}\right] \\ =\underset{x\to 0}{\lim }\left[\frac{6x}{x\left(\sqrt{1+3x}+\sqrt{1-3x}\right)}\right] \\ =\frac{6}{\sqrt{1}+\sqrt{1}} \\ =\frac{6}{2} \\ =3 \end{gathered}$$