<s>[INST] Solve $\lim_{x \to 0} \frac{\sqrt{1+2x} - \sqrt{1-4x}}{x}$ without L'Hospital's Rule. I need to solve $\lim_{x \to 0} \frac{\sqrt{1+2x} - \sqrt{1-4x}}{x}$ without  using L'Hospital's Rule. Using that rule I found the equation becomes $\lim_{x \to 0}(\frac{1}{\sqrt{1+2x}} - \frac{2}{\sqrt{1-4x}}) = \frac{-1}{\sqrt{1}}$.
However, I'm not sure how to solve this without L'Hospital's Rule. The only tool I know of is multiply  both numerator and denominator by a conjugate, but multiplying by $\frac{\sqrt{1+2x} + \sqrt{1-4x}}{\sqrt{1+2x} + \sqrt{1-4x}}$ doesn't seem to get me the same answer.
$$\lim_{x \to 0}\frac{\sqrt{1+2x} - \sqrt{1-4x}}{x} \cdot \frac{\sqrt{1+2x} + \sqrt{1-4x}}{\sqrt{1+2x} + \sqrt{1-4x}}$$
$$= \lim_{x \to 0} \frac{1+2x-(1-4x)}{x(\sqrt{1+2x} + \sqrt{1-4x})}$$
$$= \lim_{x \to 0} \frac{6x}{x(\sqrt{1+2x} + \sqrt{1-4x})}$$
$$= \lim_{x \to 0} \frac{6}{\sqrt{1+2x} + \sqrt{1-4x}}$$
$$=\frac{6}{\sqrt{1} + \sqrt{1}}$$
But $\frac{6}{\sqrt{1} + \sqrt{1}} \neq \frac{-1}{\sqrt{1}}$.
 [/INST] There is a mistake, it must be $$\frac{1}{\sqrt{1+2x}}+\frac{2}{\sqrt{1-4x}}$$
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