# L'Hospital's rule with a denominator that goes away

I am having trouble with a homework question that seems really simple, and I get the wrong answer.

$$\lim_{x\to 0} \frac{\sqrt{1+2x}-\sqrt{1-4x}}{x}$$

I then get the derivative of the top and bottom and I get

$$(1+2x)^{\frac{-1}{2}} - 2(1-4x)^\frac{-1}{2}$$

Plugging in $0$ gives me $-1$ and I am not sure why my answer doesn't match the book’s answer of $3$. I can rarely get the answer from the book so I just assume that the author does not really expect you to get the answer on your own.

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I'll +1 this question for attempting to put the question in $\TeX$.Also, I'd like if you added homework that you explicitly mention it. No worries-you'd be "guided" through. –  user21436 Mar 30 '12 at 19:39
I tried to fix it buy someone edited it first, I have no clue what was wrong. The syntax was correct and there were no open brackets. I think it has problems distinguishing brackets in brackets. –  user138246 Mar 30 '12 at 19:40
The limit of the numerator as $x\to 0$ is $1$, and the limit of the denominator is $0$, so the limit doesn’t exist. –  Brian M. Scott Mar 30 '12 at 19:41
@Jordan You may go through the edits by clicking on timestamp above Brian's name. –  user21436 Mar 30 '12 at 19:41
The limit doesn't exist. Each side limit exists in the general sense and they are $+\infty$ and $-\infty$ –  davin Mar 30 '12 at 19:41

If the book’s answer is $3$, I suspect that you’ve misquoted the problem, and that it should be $$\lim_{x\to 0}\frac{\sqrt{1+2x}-\sqrt{1-4x}}x\;.$$ This is a genuine $0/0$ indeterminate form, and after applying l’Hospital’s rule you get $$\lim_{x\to 0}\left((1+2x)^{-1/2}+2(1-4x)^{-1/2}\right)=1+2=3\;.$$

Added: And it appears that my guess was correct. You simply had a sign error in taking the derivative of the second term in the numerator.

By the way, the author does expect you to get the answer on your own. If you rarely get the answer from the book, this is an indication that either you’re missing some key ideas, or you’re making a lot of relatively minor computational errors. In this problem it was the latter.

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I see what I did now, I always mess up on math no matter how many times I double check. If a problem becomes too many steps I can never do it on my own. –  user138246 Mar 30 '12 at 19:51
@Jordan: If you have the answer in the back of the book, you can sometimes use it to see where you might have gone wrong. You had $1-2=-1$, but the correct answer was $3$. So, where did the $3$ come from? Looking at the calculation, you could guess that the $-2$ should have been a $+2$, and then you should check the derivative to see if you made an error. Obviously, this approach doesn't always work (and the answer isn't always given), but it doesn't hurt to make these kinds of observations. –  Théophile Mar 31 '12 at 15:42

L'Hospital's Rule works nicely here, but we can get by with less machinery. For note that $$\frac{\sqrt{1+2x}-\sqrt{1-4x}}{x}=\frac{(\sqrt{1+2x}-\sqrt{1-4x})(\sqrt{1+2x}+\sqrt{1-4x})}{x(\sqrt{1+2x}+\sqrt{1-4x})}.$$ When we multiply out the top, we get $(1+2x)-(1-4x)$, which is $6x$. Cancel the $x$ with the $x$ in the denominator. So we want $$\lim_{x\to 0}\frac{6}{\sqrt{1+2x}+\sqrt{1-4x}},$$ which is easy.

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I surmise that when you differentiated $-\sqrt {1\color{maroon}-4x}$, you forgot about the $\color{maroon}{\text{negative sign}}$ when using the chain rule: $${d\over dx}\textstyle\bigl(-\sqrt{1 \color{maroon}-4x}\bigr)=-{1\over2} (1-4x)^{-1/2}\cdot (\color{maroon}-4x)'=- {1\over2} (1-4x)^{-1/2}\cdot (-4)= 2(1-4x)^{-1/2}.$$

And a bit of advice, if I may:

Small mistakes can, of course, give very different answers. This is one reason why I advise that one should always write out and justify each step of the solution. Write your solution as if it would appear in the solutions manual. Among other advantages, it will be much easier then to go through your work later and discover any lurking errors.

Note that you can compute the limit as in André Nicholas' answer. (It would be good practice to go through both methods.)

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In this answer, it was shown that multiplying by $\dfrac{\sqrt{1+x}+1}{\sqrt{1+x}+1}$ yields $$\sqrt{1+x}-1=\frac{x}{\sqrt{1+x}+1}\tag{1}$$ Dividing $(1)$ by $x$ and taking limits we get $$\lim_{x\to0}\frac{\sqrt{1+x}-1}{x}=\lim_{x\to0}\frac{1}{\sqrt{1+x}+1}=\frac12\tag{2}$$ Applying $(2)$ yields \begin{align} \lim_{x\to 0}\frac{\sqrt{1+2x}-\sqrt{1-4x}}{x} &=\lim_{x\to 0}\frac{\sqrt{1+2x}-1}{x}-\lim_{x\to 0}\frac{\sqrt{1-4x}-1}{x}\\ &=2\lim_{x\to 0}\frac{\sqrt{1+2x}-1}{2x}+4\lim_{x\to 0}\frac{\sqrt{1-4x}-1}{-4x}\\ &=2\cdot\frac12+4\cdot\frac12\\ &=3\tag{3} \end{align}

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