• Support PF! Buy your school textbooks, materials and every day products Here!

A difficult limited

  • Thread starter m_s_a
  • Start date
91
0
When we think a value an end in the compound functions :
We remember that we in front of number from not a limited from the paths
 

Attachments

Answers and Replies

Dick
Science Advisor
Homework Helper
26,258
618
We remember that we in back of path from not a limited from the numbers. Is that wrong? I really can't understand you. Sorry.
 
91
0
We remember that we in back of path from not a limited from the numbers. Is that wrong? I really can't understand you. Sorry.
He is not at the end a presence
Why ؟
The previous way the others is suitable ? why ?...
Look please:
 

Attachments

Dick
Science Advisor
Homework Helper
26,258
618
He is not at the end a presence
Why ؟
The previous way the others is suitable ? why ?...
Look please:
I will. When it's approved. I think that gives me time to get a good night's sleep.
 
Dick
Science Advisor
Homework Helper
26,258
618
He is not at the end a presence
Why ؟
The previous way the others is suitable ? why ?...
Look please:
Now I don't understand what your picture is trying to tell me either. The proof of the limit in your first post is just fine.
 
91
0
Now I don't understand what your picture is trying to tell me either. The proof of the limit in your first post is just fine.

I did not understand your intention ?
I do not have a confirmation of the response
But look at the next example:
I apologize
Second : on the axis z = y^2
 

Attachments

Last edited:
HallsofIvy
Science Advisor
Homework Helper
41,732
893
You have now posted 3 different pictures that appear to involve three different problems. As Dick told you before, the first is correct. This third one is also correct- except for spelling: since you get two different limits by approaching (0,0) along two different paths, the limit does not exist.

I'm not sure what the second picture was supposed to represent. But one thing you should be careful about: The line y= 2x, for example, in R2, is not in the complex plane.
 
91
0
You have now posted 3 different pictures that appear to involve three different problems. As Dick told you before, the first is correct. This third one is also correct- except for spelling: since you get two different limits by approaching (0,0) along two different paths, the limit does not exist.
My answer of the first question :
is correct or a mistake?


I'm not sure what the second picture was supposed to represent. But one thing you should be careful about: The line y= 2x, for example, in R2, is not in the complex plane.

My answer of the first question :
is correct or a mistake?


remember:
x=(z+z')/2
y=(z-z')/2i
z'=z par
 

Attachments

Dick
Science Advisor
Homework Helper
26,258
618
lim(z->0) is not the same as lim(x->0) or lim(y->0). It's lim(x->0 AND y->0).
 
91
0
lim(z->0) is not the same as lim(x->0) or lim(y->0). It's lim(x->0 AND y->0).
Look into the proof of Coshi theory - Riman

f'(z)=f'(x)=f'(y)


no f'(z)=f'(x)+f'(y)
 
Dick
Science Advisor
Homework Helper
26,258
618
Look into the proof of Coshi theory - Riman

f'(z)=f'(x)=f'(y)


no f'(z)=f'(x)+f'(y)
What have derivatives got to do with it? You were posting limit problems. Or something. I give up. I don't know what your question or problem is.
 
91
0
I support the second solution
And I am not sure
 
Dick
Science Advisor
Homework Helper
26,258
618
If you are asking does |z|^2/z approach 0 as z approaches 0. Then, it does. You're first argument is fine. If you think there is a problem because the limit of z* as z approaches 0 is path dependent, then you are wrong. The limit is 0.
 

Related Threads for: A difficult limited

  • Last Post
Replies
2
Views
690
  • Last Post
Replies
7
Views
1K
  • Last Post
Replies
14
Views
581
  • Last Post
Replies
2
Views
1K
  • Last Post
Replies
10
Views
1K
  • Last Post
Replies
2
Views
1K
Replies
6
Views
4K
Replies
4
Views
686
Top