Find the value of the one sided limit

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

Thank you very much for all your help!

Printable View

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

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, 04:20 AMTriKri
Since both the numerator and the denominator , use l’Hopital’s rule.

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

since the denominator is undefined

etc...

- Adam - December 5th 2006, 05:04 AMTriKri

- December 5th 2006, 05:58 AMThePerfectHacker
Factor, your idea was correct.

Now when we approach the limit from the right,

that means,

But then,

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,

on some open interval where (a really small number). Because we do not care what happens far away one what happens really close to the point.

In that case,

Thus, the radical is defined.

And furthermore we can factor,

- December 5th 2006, 09:46 AMtopsquark
- December 5th 2006, 10:07 AMThePerfectHacker
- December 5th 2006, 10:15 AMputnam120
just because the denominator tends to zero does not mean that you can use l'hopital.

for instance

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, 12:35 PMThePerfectHacker
- December 5th 2006, 01:03 PMTriKri
- December 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 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, 04:56 PMThePerfectHacker
- December 6th 2006, 04:13 AMTriKri
But why doesn't the right hand exist in this case? Isn't complex numbers allowed?

- December 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 (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, 01:50 PMTriKri