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Please, Give us something readable without us having to switch back & forth between tabs/windows.
I wasn't 100% sure of how proving that limit equals zero would help but I did take your advice and broke the problem into several limits whose products are infinity.Instead of trying to bound the expression above by 1/N, why not apply what you know about the algebra of limits?
[tex]\lim_{z \rightarrow i} \frac{(z + i)(z^2 - 1)(z - i)}{z^2 + i} = \frac{[\lim_{z \rightarrow i} (z + i)] [\lim_{z \rightarrow i} (z^2 - 1)][\lim_{z \rightarrow i} (z - i)]}{\lim_{z \rightarrow i} (z^2 + i)}[/tex]
This is valid provided that the limit in the denominator is nonzero.
If you can prove that the above limit is zero, then that implies your 1/N bound and you can proceed accordingly.
I used the idea of bounding |z+i| by 3 by assuming delta was at most 1. I attached a pdf of my work. If someone could check it to make sure my logic is alright I'd be very grateful. I've just recently found out how epsilon and deltas work.thed0ctor,
See what bound [itex]\displaystyle \left|z-i\right|<\frac{1}{2}[/itex] puts on [itex]\displaystyle \left|\frac{(z + i)(z^2 - 1)}{z^2 + i}\right|\ .[/itex]
[itex]\displaystyle \frac{-1+i}{\sqrt{2}}[/itex] is less that 1 unit from i on the complex plane, so δ = 1 is too large....
I used the idea of bounding |z+i| by 3 by assuming delta was at most 1. I attached a pdf of my work. If someone could check it to make sure my logic is alright I'd be very grateful. I've just recently found out how epsilon and deltas work.
Sorry the last bit was meant to say[itex]\displaystyle \frac{-1+i}{\sqrt{2}}[/itex] is less that 1 unit from i on the complex plane, so δ = 1 is too large.
[itex]\displaystyle \left(\frac{-1+i}{\sqrt{2}}\right)^2=-i[/itex] giving division by zero for the denominator of z^{2} + i .
That depends ...Sorry the last bit was meant to say
[tex]∴ \lim_{z \rightarrow i} \frac{z^2 + i}{z^4 -1} = ∞[/tex]
not:
[tex]∴ \lim_{z \rightarrow i} \frac{z^2 + i}{z^4 -i} = ∞[/tex]
Does this fix that issue? Because now I don't see the division by zero issue.
Good point. I also believe I found a bound. it's a nasty number but it's off the assumption that |z-i|< δ< .5That depends ...
Is it acceptable to split-up the limit as you did in the PDF file? If so, then the proof at the end is fine.
However, if you are to use the definition for the given expression, then your first attempt was a good start. The algebra may be pretty involved to find what the bound is, depending on what restriction you place on δ.