# 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.