Does the Limit of (sqrt(|ab|) * |h|) / h as h Approaches 0 Exist?

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The discussion revolves around the limit of (sqrt(|ab|) * |h|) / h as h approaches 0. Participants clarify that as h approaches 0, the expression does not converge to a single value, resulting in a limit of +/- sqrt(|ab|). The confusion stems from the interpretation of sqrt(|ab|) as a constant, which does not change with h. The conclusion is that while the partial derivatives exist, the directional derivatives in other directions do not, confirming that the limit does not exist. The final agreement is that the limit as h approaches 0 of sqrt(|ab|) is indeed sqrt(|ab|), independent of h.
Callisto
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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
 
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Why doesn't √|a*b| go to √|a*b| as h goes to ∞?
 
as h goes to infinity, sqrt|(a*b)| goes to zero
 
Why?Is there something that u're hiding,like a dependence of "a" & "b" of "h"...?:bugeye:


Daniel.
 
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?
please forgive my confusion.

Callisto
 
Are a and b constants?
 
as long as a&b do not equal zero
 
What is \lim_{h \rightarrow \infty} 1?
 
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 guarantee that sqrt|(a*b)| is greater than 100? What if I made N=1,000,000 instead then what would d be?
 
  • #10
You're confusing me.

is this the answer to my question?

callisto
 
  • #11
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
Callisto said:
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

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
<br /> \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}

:wink:
 
  • #14
master_coda said:
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.

You might want to check that.
 
  • #15
ONly if you have failed to disclose what a and b are. you were asked but haven't bothered to say.
 
  • #16
Callisto said:
as long as a&b do not equal zero

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
And that just 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 just fix a and b. sqrt(ab) tends to sqrt(ab) as h tends to zero as it is independent of it.
 
  • #18
Callisto said:
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

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 don't know how to make it anymore clearer.

callisto :!)
 
  • #19
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
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.
 
  • #21
Callisto said:
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.

Yes, that is correct.
 
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