Answer
$3$
Work Step by Step
Given: $\lim\limits_{x\to0}\dfrac{\sqrt{1+2x}-\sqrt{1-4x}}{x}$
Need to apply L'Hospital's Rule because this shows an indeterminate form of $\dfrac{0}{0}$.
Let us multiply both numerator and denominator by $\sqrt{1+2x}+\sqrt{1-4x}$.
Then $\sqrt{1+2x}-\sqrt{1-4x})(\sqrt{1+2x}-\sqrt{1-4x})=(1+2x)-(1-4x)=6x$
and
$\lim\limits_{x\to0}\dfrac{6x}{x(\sqrt{1+2x}+\sqrt{1-4x})}=\lim\limits_{x\to0}\frac{6}{\sqrt{1+2x}+\sqrt{1-4x}}$
or, $=\dfrac{6}{\sqrt{1+2\times0}+\sqrt{1-4\times0}}$
or, $=\dfrac{6}{\sqrt 1+\sqrt 1}$
or, $=\dfrac{6}{2}$
or, $=3$