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

  • Thread starter Callisto
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In summary: The limit of sqrt(|ab|) as h tends to zero is sqrt(|ab|) because sqrt(|ab|) is independent of h. This means that the directional derivatives in all other directions fail to exist. In summary, the limit of (sqrt|(a*b)|*|h|)/h as h tends to zero is +/- sqrt|a*b|, and the directional derivatives in all other directions do not exist as sqrt|a*b| is independent of h.
  • #1
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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  • #2
Why doesn't √|a*b| go to √|a*b| as h goes to ∞?
 
  • #3
as h goes to infinity, sqrt|(a*b)| goes to zero
 
  • #4
Why?Is there something that u're hiding,like a dependence of "a" & "b" of "h"...?:bugeye:


Daniel.
 
  • #5
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
 
  • #6
Are a and b constants?
 
  • #7
as long as a&b do not equal zero
 
  • #8
What is [itex]\lim_{h \rightarrow \infty} 1[/itex]?
 
  • #9
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
[tex]
\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}[/tex]

: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.
 

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

1. What is a limit that does not exist?

A limit that does not exist refers to a mathematical concept where the value of a function at a certain point does not approach a specific value as the input approaches that point. This can happen when there is a jump or discontinuity in the function at that point.

2. How can you determine if a limit does not exist?

A limit does not exist if the left-hand limit and the right-hand limit at a specific point do not equal each other. This means that the function is not continuous at that point.

3. Can a limit not exist at multiple points?

Yes, a limit can not exist at multiple points on the same function. This can happen when there are multiple jumps or discontinuities within the function.

4. What is the difference between a limit not existing and a limit being infinite?

A limit not existing means that the function does not approach a specific value as the input approaches a certain point. On the other hand, a limit being infinite means that the function approaches either positive or negative infinity as the input approaches a certain point.

5. Can a limit not exist if the function is defined at that point?

Yes, a function can be defined at a certain point but still have a limit that does not exist. This can happen when there is a removable discontinuity at that point, meaning that the function can be made continuous by filling in the gap.

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