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- Thread starter m_s_a
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Dick

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He is not at the end a presence

Why ؟

The previous way the others is suitable ? why ?...

Look please:

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Dick

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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.

- #5

Dick

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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.

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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

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- #7

HallsofIvy

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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

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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 thecomplexplane.

My answer of the first question :

is correct or a mistake?

remember:

x=(z+z')/2

y=(z-z')/2i

z'=z par

- #9

Dick

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lim(z->0) is not the same as lim(x->0) or lim(y->0). It's lim(x->0 AND y->0).

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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)

- #11

Dick

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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.

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I support the second solution

And I am not sure

And I am not sure

- #13

Dick

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