How to find $\;\lim_{x\to 0} \frac{x}{2-\sqrt{4}-2x}\;$ without using L'Hôpital's rule? Well, I'm studying calculus and doing some exercises. First of all, from the answers that were given by my teacher the result of this limit should be $4$. I'm beginning my calculus class now but I've already studied calculus by myself before so I know how to apply l'Hôpital's Rule to this limit, following l'Hôpital's I get $-1$, the step-by-step resolution answer from WolframAlpha gives me $-1$ as well. 
I thought of all I know about factoring and simplifying equations but I'm not able to actually solve it without l'Hôpital's. Any hint or way about how to do it?
$$\lim\limits_{x\to 0}  \frac{x}{2-\sqrt{4}-2x}$$
 A: $$\lim\limits_{x\to 0}  \frac{x}{\underbrace{2-\sqrt{4}}_{2 - 2 = 0}-2x} = \lim_{x \to 0} \frac x{-2x} = -\frac 12$$
A: Note: The function in the OP has gone through several edits, and is now not the function discussed below. 
Hint: Multiply top and bottom by $2+\sqrt{4-2x}$. Very nice things happen. 
Remark: An interesting alternative is to consider the reciprocal. We will also change notation a bit, and look for
$$\lim_{h\to 0}-\frac{\sqrt{4-2h}-2}{h}.\tag{$1$}$$
We recognize 
$$\lim_{h\to 0}\frac{\sqrt{4-2h}-2}{h}.$$
as the derivative of $f(x)=\sqrt{4-2x}$ at $x=0$. If we are already familiar with differentiation "rules," we can compute the derivative of $\sqrt{4-2x}$, set $x=0$, and insert the minus sign from $(1)$. Now take the reciprocal to get the answer to the original problem.  
Note that if the function is indeed the one we have calculated with, the limit is neither $4$ nor $-1$. 
A: It is fairly simple to solve the question. It happens to the best of us but it seems that you have not noticed that you can simplify the denominator $2-\sqrt{4}-2x$ to $-2x$. So then you will have: 
$$\lim\limits_{x\to 0}  \frac{x}{-2x}$$
Which then simplifies to
$$\lim\limits_{x\to 0}  \frac{1}{-2}$$
So your limit is: 
$$\lim\limits_{x\to 0}  \frac{x}{2-\sqrt{4}-2x} = -\frac{1}{2}$$
