Homework Help: A difficult limited

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.

 

I will. When it's approved. I think that gives me time to get a good night's sleep.

 

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.

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.

 

lim(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.

 

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.