TITLE: L'Hospital's rule with a denominator that goes away
QUESTION [6 upvotes]: 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.

REPLY [3 votes]: 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
$$
\begin{align}
\lim_{x\to 0}\frac{\sqrt{1+2x}-\sqrt{1-4x}}{x}
&=\lim_{x\to 0}\frac{\sqrt{1+2x}-1}{x}-\lim_{x\to 0}\frac{\sqrt{1-4x}-1}{x}\\
&=2\lim_{x\to 0}\frac{\sqrt{1+2x}-1}{2x}+4\lim_{x\to 0}\frac{\sqrt{1-4x}-1}{-4x}\\
&=2\cdot\frac12+4\cdot\frac12\\
&=3\tag{3}
\end{align}
$$