Does the Limit in the Complex Plane Approach Infinity?

cragar
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Homework Statement


lim as z--> i , \frac{z^2-1}{z^2+1}

The Attempt at a Solution


[/B]When we plug in i we get -2/0, so we get division by 0, Does this mean the limit is
infinity, I also tried approaching from z=x+i where x went to 0, you get the same answer,
I also approached from z=yi where y approaches 1, and I got the same answer.
 
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cragar said:

Homework Statement


lim as z--> i , \frac{z^2-1}{z^2+1}

The Attempt at a Solution


[/B]When we plug in i we get -2/0, so we get division by 0, Does this mean the limit is
infinity, I also tried approaching from z=x+i where x went to 0, you get the same answer,
I also approached from z=yi where y approaches 1, and I got the same answer.
What happens with ##\lim_{x \to 0} \frac 1 x##? Is the limit here ##\infty## or does the limit simply not exist at all? IOW, are the two one-sided limits equal?
 
good point if we appraoch x from the left it goes to minus infinity , and if we approach from the right it is positive infinity, so the limit does not exist.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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