# One sided limit

• December 5th 2006, 04:31 AM
One sided limit
Find the value of the one sided limit
$\lim_{x \to 2} \frac{4-x^2}{\sqrt{2+x-x^2}}$

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

Thank you very much for all your help!
• December 5th 2006, 05:20 AM
TriKri
Quote:

Find the value of the one sided limit
$\lim_{x \to 2} \frac{4-x^2}{\sqrt{2+x-x^2}}$

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

Thank you very much for all your help!

Since both the numerator and the denominator $\longrightarrow 0$, use l’Hopital’s rule.
• December 5th 2006, 05:53 AM
I got it now
i got it now
since the denominator is undefined
$\lim_{x \to 2^-} \frac{(2-x)(2+x)}{\sqrt {(2-x)(1+x)}}$
etc...

• December 5th 2006, 06:04 AM
TriKri
$\frac{(2-x)(2+x)}{\sqrt {(2-x)(1+x)}}\ =$

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

$\sqrt {0\cdot (1+x)}\ =\ 0$
• December 5th 2006, 06:58 AM
ThePerfectHacker
Quote:

Find the value of the one sided limit
$\lim_{x \to 2} \frac{4-x^2}{\sqrt{2+x-x^2}}$

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

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

Now when we approach the limit from the right,
$x>2$ that means,
$2-x<0$
$1+x>0$
But then,
$(2-x)(1+x)<0$
And the function is undefined.

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,
$x<2$ on some open interval $(\delta, 2)$ where $\delta <2$ (a really small number). Because we do not care what happens far away one what happens really close to the point.
In that case,
$2-x>0$
$1+x>0$
And furthermore we can factor,
$\lim_{x\to 2^-} \frac{(2-x)(2+x)}{\sqrt{2-x}\cdot \sqrt{1+x}}$
$\lim_{x\to 2^-} \frac{\sqrt{2-x}(2+x)}{\sqrt{1+x}}=\frac{0\cdot 4}{3}=0$
• December 5th 2006, 10:46 AM
topsquark
Quote:

Originally Posted by TriKri
Since both the numerator and the denominator $\longrightarrow 0$, use l’Hopital’s rule.

But since the denominator is not continuous at x = 2 CAN we use L'Hopital?

-Dan
• December 5th 2006, 11:07 AM
ThePerfectHacker
Quote:

Originally Posted by topsquark
But since the denominator is not continuous at x = 2 CAN we use L'Hopital?

-Dan

Yes you can on the half interval.

In fact, the proof of the L'Hopital theorem is done in two parts. Left and Right hand limits.
• December 5th 2006, 11:15 AM
putnam120
just because the denominator tends to zero does not mean that you can use l'hopital.

for instance

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

it is obvious that the limit does not exist.
in order to use l'hopital u need to have an indeterminate form.
• December 5th 2006, 01:35 PM
ThePerfectHacker
Quote:

Originally Posted by putnam120
just because the denominator tends to zero does not mean that you can use l'hopital.

for instance

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

it is obvious that the limit does not exist.
in order to use l'hopital u need to have an indeterminate form.

I think he was asking whether you can use it here. In this example. Because the numerator is not continous on an open interval conting 1, rather on a half-open interval.
• December 5th 2006, 02:03 PM
TriKri
Quote:

Originally Posted by topsquark
But since the denominator is not continuous at x = 2 CAN we use L'Hopital?

-Dan

Well, I don't know, but I think so :o You just have to choose whether x > 2 or x < 2 I guess ... do you have an example of when it doesn't work?
• December 5th 2006, 02:47 PM
topsquark
Quote:

Originally Posted by TriKri
Well, I don't know, but I think so :o You just have to choose whether x > 2 or x < 2 I guess ... do you have an example of when it doesn't work?

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 $\sqrt{2 + x - x^2}$ 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
• December 5th 2006, 05:56 PM
ThePerfectHacker
Quote:

Originally Posted by topsquark
Apparently as long as the limit of the derivative exists, L'Hopital is usable, if I'm understanding TPH correctly.

As long as the "left handed" or "right handed" derivates exist.
• December 6th 2006, 05:13 AM
TriKri
But why doesn't the right hand exist in this case? Isn't complex numbers allowed?
• December 6th 2006, 05:18 AM
ThePerfectHacker
Quote:

Originally Posted by TriKri
Isn't complex numbers allowed?

First I would like to mention square roots of of negative numbers are undefined. Note, it is mathematically incorrect to say $\sqrt{-4}=2i$ (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.
• December 6th 2006, 02:50 PM
TriKri
Quote:

Originally Posted by ThePerfectHacker
First I would like to mention square roots of of negative numbers are undefined. Note, it is mathematically incorrect to say $\sqrt{-4}=2i$ (I know that is what they teach in high schools, but it is still wrong).

My world picture has been shattered! :( ;_;