# 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?

$$\lim\limits_{x\to 0} \frac{x}{2-\sqrt{4}-2x}$$

• en.wikipedia.org/wiki/Conjugate_(algebra) – Double AA May 23 '13 at 1:12
• $\sqrt{4}=2$ Why not just make the denomenator $-2x$? – Double AA May 23 '13 at 1:14
• Yes @DoubleAA, that is what I meant, just typed like the example shows, but yeah, it's 2. – Thums May 23 '13 at 1:16
• This is weird. If the problem is indeed as written in the question, the limit is $-\frac12$. If the square root is meant to go over $4-2x$, the limit is $2$. What are you typing into Wolfram Alpha to get $-1$? – Javier May 23 '13 at 1:21
• @LuanCristianThums But Wolfram and l'Hôpital should yield $-\frac{1}{2}$ – Double AA May 23 '13 at 1:21

$$\lim\limits_{x\to 0} \frac{x}{\underbrace{2-\sqrt{4}}_{2 - 2 = 0}-2x} = \lim_{x \to 0} \frac x{-2x} = -\frac 12$$

• Very slow day for me too! +1 – Amzoti May 23 '13 at 1:26
• What in the World could dare to annoy you such that? – mrs May 23 '13 at 18:53
• Dear Luan: Did this help? – Namaste May 24 '13 at 2:37

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 $-1$.

• Please see the edited question. – Thums May 23 '13 at 1:21
• Strange denominator! $2-\sqrt{4}-2x=-2x$. – André Nicolas May 23 '13 at 1:34
• Thank you, @amWhy and DoubleAAA. The thing is that the question is incorrectly typed and I incorrectly typed the incorrect question here, what a confusion. Anyway, your answers helped me a lot, thanks! And sorry for my bad english. – Thums May 23 '13 at 1:36

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: $$\lim\limits_{x\to 0} \frac{x}{2-\sqrt{4}-2x} = -\frac{1}{2}$$