 #1
 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
We remember that we in front of number from not a limited from the paths
Attachments

173.2 KB Views: 445
He is not at the end a presenceWe remember that we in back of path from not a limited from the numbers. Is that wrong? I really can't understand you. Sorry.
I will. When it's approved. I think that gives me time to get a good night's sleep.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.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.
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 R^{2}, is not in the complex plane.
Look into the proof of Coshi theory  Rimanlim(z>0) is not the same as lim(x>0) or lim(y>0). It's lim(x>0 AND y>0).
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.Look into the proof of Coshi theory  Riman
f'(z)=f'(x)=f'(y)
no f'(z)=f'(x)+f'(y)