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?

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

  • $\begingroup$ en.wikipedia.org/wiki/Conjugate_(algebra) $\endgroup$ – Double AA May 23 '13 at 1:12
  • $\begingroup$ $\sqrt{4}=2$ Why not just make the denomenator $-2x$? $\endgroup$ – Double AA May 23 '13 at 1:14
  • $\begingroup$ Yes @DoubleAA, that is what I meant, just typed like the example shows, but yeah, it's 2. $\endgroup$ – Thums May 23 '13 at 1:16
  • $\begingroup$ 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 $-1$? $\endgroup$ – Javier May 23 '13 at 1:21
  • $\begingroup$ @LuanCristianThums But Wolfram and l'Hôpital should yield $-\frac{1}{2}$ $\endgroup$ – Double AA May 23 '13 at 1:21

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

  • $\begingroup$ Very slow day for me too! +1 $\endgroup$ – Amzoti May 23 '13 at 1:26
  • $\begingroup$ What in the World could dare to annoy you such that? $\endgroup$ – mrs May 23 '13 at 18:53
  • $\begingroup$ Dear Luan: Did this help? $\endgroup$ – 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 $-1$.

  • $\begingroup$ Please see the edited question. $\endgroup$ – Thums May 23 '13 at 1:21
  • $\begingroup$ Strange denominator! $2-\sqrt{4}-2x=-2x$. $\endgroup$ – André Nicolas May 23 '13 at 1:34
  • $\begingroup$ 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. $\endgroup$ – 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: $$\lim\limits_{x\to 0} \frac{x}{2-\sqrt{4}-2x} = -\frac{1}{2}$$


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