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.
 A: 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.
A: 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.
A: 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.)
A: 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
$$
\begin{align}
\lim_{x\to 0}\frac{\sqrt{1+2x}-\sqrt{1-4x}}{x}
&=\lim_{x\to 0}\frac{\sqrt{1+2x}-1}{x}-\lim_{x\to 0}\frac{\sqrt{1-4x}-1}{x}\\
&=2\lim_{x\to 0}\frac{\sqrt{1+2x}-1}{2x}+4\lim_{x\to 0}\frac{\sqrt{1-4x}-1}{-4x}\\
&=2\cdot\frac12+4\cdot\frac12\\
&=3\tag{3}
\end{align}
$$
