Answer
$$\lim_{x\to0}\frac{\sqrt{1+2x}-\sqrt{1-4x}}{x}=3$$
Work Step by Step
$$A=\lim_{x\to0}\frac{\sqrt{1+2x}-\sqrt{1-4x}}{x}$$
L'Hospital's Rule can be applied in this exercise since this is an indeterminate form of $\frac{0}{0}$. However, to take the derivative of $\sqrt{1+2x}-\sqrt{1-4x}$ and then transform and simplify it to the ultimate result may be a little bit tiresome. Therefore, it might be quicker to stick to the elementary way.
We multiply both numerator and denominator by $\sqrt{1+2x}+\sqrt{1-4x}$.
The numerator hence would be
$$(\sqrt{1+2x}-\sqrt{1-4x})(\sqrt{1+2x}-\sqrt{1-4x})$$
$$=(1+2x)-(1-4x)$$
$$=6x$$
Thus, $A$ would now be
$$A=\lim_{x\to0}\frac{6x}{x(\sqrt{1+2x}+\sqrt{1-4x})}$$
$$A=\lim_{x\to0}\frac{6}{\sqrt{1+2x}+\sqrt{1-4x}}$$
$$A=\frac{6}{\sqrt{1+2\times0}+\sqrt{1-4\times0}}$$
$$A=\frac{6}{\sqrt 1+\sqrt 1}$$
$$A=\frac{6}{2}=3$$
