Find the value of the one sided limit

$\displaystyle \lim_{x \to 2} \frac{4-x^2}{\sqrt{2+x-x^2}}$

WHere will the lim be shifting from the left( - ) or from the right ( + )? and why?

Thank you very much for all your help!

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- Dec 5th 2006, 03:31 AM^_^Engineer_Adam^_^One sided limit
Find the value of the one sided limit

$\displaystyle \lim_{x \to 2} \frac{4-x^2}{\sqrt{2+x-x^2}}$

WHere will the lim be shifting from the left( - ) or from the right ( + )? and why?

Thank you very much for all your help! - Dec 5th 2006, 04:20 AMTriKri
Since both the numerator and the denominator $\displaystyle \longrightarrow 0$, use l’Hopital’s rule.

- Dec 5th 2006, 04:53 AM^_^Engineer_Adam^_^I got it now
i got it now

since the denominator is undefined

$\displaystyle \lim_{x \to 2^-} \frac{(2-x)(2+x)}{\sqrt {(2-x)(1+x)}}$

etc...

- Adam - Dec 5th 2006, 05:04 AMTriKri
$\displaystyle \frac{(2-x)(2+x)}{\sqrt {(2-x)(1+x)}}\ =$

$\displaystyle \sqrt {(2-x)(1+x)}\ \to \left/\lim_{x \to 0}\right/ \to$

$\displaystyle \sqrt {0\cdot (1+x)}\ =\ 0$ - Dec 5th 2006, 05:58 AMThePerfectHacker
Factor, your idea was correct.

$\displaystyle \lim_{x\to 2} \frac{(2-x)(2+x)}{\sqrt{(2-x)(1+x)}}$

Now when we approach the limit from the right,

$\displaystyle x>2$ that means,

$\displaystyle 2-x<0$

$\displaystyle 1+x>0$

But then,

$\displaystyle (2-x)(1+x)<0$

And the function is undefined.

(Radical is a negative number!)

So the limit (the regular limit)**does not exist**.

However, the one-handed limit, might. We already shown that the right-handed does not exists. Let us see the left-handed.

When we approach the limit from the left,

$\displaystyle x<2$ on some open interval $\displaystyle (\delta, 2)$ where $\displaystyle \delta <2$ (a really small number). Because we do not care what happens far away one what happens really close to the point.

In that case,

$\displaystyle 2-x>0$

$\displaystyle 1+x>0$

Thus, the radical is defined.

And furthermore we can factor,

$\displaystyle \lim_{x\to 2^-} \frac{(2-x)(2+x)}{\sqrt{2-x}\cdot \sqrt{1+x}}$

$\displaystyle \lim_{x\to 2^-} \frac{\sqrt{2-x}(2+x)}{\sqrt{1+x}}=\frac{0\cdot 4}{3}=0$ - Dec 5th 2006, 09:46 AMtopsquark
- Dec 5th 2006, 10:07 AMThePerfectHacker
- Dec 5th 2006, 10:15 AMputnam120
just because the denominator tends to zero does not mean that you can use l'hopital.

for instance

$\displaystyle \lim_{x\to 1}\frac{1}{1-x}$

it is obvious that the limit does not exist.

in order to use l'hopital u need to have an indeterminate form. - Dec 5th 2006, 12:35 PMThePerfectHacker
- Dec 5th 2006, 01:03 PMTriKri
- Dec 5th 2006, 01:47 PMtopsquark
I don't have an example of where it doesn't work, I was simply speculating. I was just trying to be careful since the derivative of $\displaystyle \sqrt{2 + x - x^2}$ doesn't exist at x = 2 (because the function doesn't exist for x < 2.) Apparently as long as the limit of the derivative exists, L'Hopital is usable, if I'm understanding TPH correctly.

-Dan - Dec 5th 2006, 04:56 PMThePerfectHacker
- Dec 6th 2006, 04:13 AMTriKri
But why doesn't the right hand exist in this case? Isn't complex numbers allowed?

- Dec 6th 2006, 04:18 AMThePerfectHacker
First I would like to mention square roots of of negative numbers are

**undefined**. Note, it is mathematically incorrect to say $\displaystyle \sqrt{-4}=2i$ (I know that is what they teach in high schools, but it is still wrong). I have already discussed that on the forum. Next, all the L'Hopitals rule was only proven for*real*differenciable functions (at least I think). Finally this problem it self is*real*, implies that the function the poster was using mapped the real numbers into the real numbers. - Dec 6th 2006, 01:50 PMTriKri