Limits that do not exist?

1. Apr 12, 2005

Callisto

Limits that do not exist??

How do i show that

lim h->0 (sqrt|(a*b)|*|h|)/h does not exist at 0

so far i have
lim h->0+ |h|/h =1
and
lim h->0- |h|/h =-1

does sqrt|(a*b)| go to +/- inifinity?
because as h gets smaller, sqrt|(a*b)| goes infinity.

so
lim h->0 (sqrt|(a*b)|*|h|)/h = +/- inifinity
Callisto

2. Apr 12, 2005

Hurkyl

Staff Emeritus

3. Apr 12, 2005

Callisto

as h goes to infinity, sqrt|(a*b)| goes to zero

4. Apr 12, 2005

dextercioby

Why?Is there something that u're hiding,like a dependence of "a" & "b" of "h"...?

Daniel.

5. Apr 12, 2005

Callisto

OK my question is,

does sqrt|(a*b)| go to +/- inifinity?
because as h gets smaller, sqrt|(a*b)| goes to infinity?

I thought that as h->0 then sqrt|(a*b)| -> infinity.
am i not computing the limit correctly?

Callisto

6. Apr 12, 2005

Moo Of Doom

Are a and b constants?

7. Apr 12, 2005

Callisto

as long as a&b do not equal zero

8. Apr 12, 2005

Hurkyl

Staff Emeritus
What is $\lim_{h \rightarrow \infty} 1$?

9. Apr 12, 2005

snoble

I think perhaps you may not understand the terms you are using.
When you say
"as h->0 then sqrt|(a*b)| -> infinity"
That means given any number N>0 there exists an d>0 such that if |h|<d then sqrt|(a*b)|>N.

For example |1/h| -> infinity as h -> 0 because given an N>0 I can
take d= 1/2N so if 0<|h|<d then |1/h| > 1/d =2N >N.

So can you do that with sqrt|(a*b)|? If I gave you N=100 what would you give me for d? Why for that d that you've given me can you guarentee that sqrt|(a*b)| is greater than 100? What if I made N=1,000,000 instead then what would d be?

10. Apr 12, 2005

Callisto

You're confusing me.

is this the answer to my question?

callisto

11. Apr 13, 2005

Callisto

If f(x,y) = sqrt|a*b|, the partial derivatives at Dx(0,0) ,Dy(0,0) exist. i am trying to show that the directional derivatives in all other directions fail to exist.

Callisto

12. Apr 13, 2005

master_coda

If f(x,y) = sqrt|a*b|, then the directional derivatives also exist. They are all 0. Because sqrt|a*b| is a constant. It doesn't change when x or y do.

And if you take a limit as h -> 0, then sqrt|a*b| -> sqrt|a*b|. Because sqrt|a*b| is a constant. It doesn't change when h does.

13. Apr 13, 2005

Zurtex

$$\lim_{\substack{a\rightarrow \infty\\b\rightarrow \infty\\c\rightarrow \infty\\x\rightarrow \infty\\y\rightarrow \infty\\z\rightarrow \infty\\\alpha\rightarrow \infty\\\beta\rightarrow \infty\\\gamma\rightarrow \infty\\\delta\rightarrow \infty}} \sqrt{4 \cdot 3} = \sqrt{4 \cdot 3}$$

14. Apr 13, 2005

Callisto

You might want to check that.

15. Apr 13, 2005

matt grime

ONly if you have failed to disclose what a and b are. you were asked but haven't bothered to say.

16. Apr 13, 2005

Callisto

Means a does not equal zero and b does not equal zero.

From the definition of the directional derivative

ie: The directional derivative of f at (x,y) in the direction of the unit vector
u=<a,b> is

Duf(x,y)=lim h->0 f(x+ha,y+hb)-f(x,y)/h

IF THIS LIMIT EXISTS.

17. Apr 13, 2005

matt grime

And that jsut confirms that you're confused abtou these things. partly a problem of the notation.

for instance, writing f(x,y) = sqrt(ab) implies a and b are constants, so that taking partial differentials kills it.

anyway, the poitn was jsut fix a and b. sqrt(ab) tends to sqrt(ab) as h tends to zero as it is independent of it.

18. Apr 13, 2005

Callisto

The confusion has only arisen as a result of the misinterpretation of my question on you and your colleagues behalf.

since the sqrt|a*b| does not equal zero,

and the limit of |h|/h = 1 as h tends to zero from the right and the limit of |h|/h=-1 as h tends to zero from the left.

then the limit (sqrt|a*b|*|h|)/h as h tends to zero is +/- sqrt|a*b|.

therefore the directional derivatives in all other directions fail to exist

I must add i am not a mathematician so i dont know how to make it anymore clearer.

callisto :!!)

19. Apr 13, 2005

matt grime

We are correctly intepreting what you wrote in and of itself. However what you wrote was confused.

Even the quote you use from yourself implies that f is a constant function, when it isn't.

But the very first post contained some errors, that have been corrected. You do understand that the limt as h tends to zero of sqrt(|ab|) is sqrt(|ab|)?

HOwever, you have got the gitst of why the derivatives do not exist.

20. Apr 13, 2005

Callisto

So am i correct?

the limit as h tends to zero of sqrt(|ab|) is sqrt(|ab|) since sqrt(|ab|) is independent of h

I am an undergraduate forgive me for the confusion.