# 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 $$\frac{6}{\sqrt{1} + \sqrt{1}} \neq \frac{-1}{\sqrt{1}}$$.

• Your answer is actually correct : $\frac 6{\sqrt 1 + \sqrt 1} = 3$ is correct. This means there is something wrong with how you got $\frac {-1}{\sqrt 1} = -1$, so you have made some error while doing L'Hopital it seems! Rationalizing is certainly the first thought coming to mind if you aren't doing L'Hopital in this case, so "the only tool you know" is good enough here There are not too many other rules around, really, at the calc 1 level that you need to know. – Teresa Lisbon Apr 5 '19 at 16:34
• @астонвіллаолофмэллбэрг Oh! okay, that changes things. I'll take a look at how I got $\frac{-1}{\sqrt{1}}$ – LuminousNutria Apr 5 '19 at 16:40
• Yes. Post that on the question if you cannot figure out your mistake, we'll do it together! – Teresa Lisbon Apr 5 '19 at 16:41
• @астонвіллаолофмэллбэрг I got it. My mistake was that I thought $\frac{d}{dx}\sqrt{1-4x} = \frac{2}{\sqrt{1-4x}}$ but it was really $\frac{-2}{\sqrt{1-4x}}$. – LuminousNutria Apr 5 '19 at 16:46
• All right good to know that it's been solved – Teresa Lisbon Apr 5 '19 at 18:32

$$\frac{\sqrt{1+2x}-\sqrt{1-4x}}{x} =\frac{1+x + o(x)-1 + 2x + o(x)}{x} = 3 + o(1)$$
So the limit is $$3$$.
• I am only a calc 1 student, and I'm not really advanced enough to understand what you're doing here. Could you please explain it a little more specifically? I don't know what the $o(x)$ and $o(1)$ notation represent. – LuminousNutria Apr 5 '19 at 16:35
There is a mistake, it must be $$\frac{1}{\sqrt{1+2x}}+\frac{2}{\sqrt{1-4x}}$$