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thed0ctor
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Please, Give us something readable without us having to switch back & forth between tabs/windows.thed0ctor said:
jbunniii said: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.
SammyS said: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.thed0ctor said:...
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.
SammyS said:[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 z2 + i .
That depends ...thed0ctor said: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.
SammyS said:That 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.
"lim" stands for limit and "z->i" means that the variable z is approaching the value of i. This notation is used to represent the behavior of a function as its input approaches a specific value.
To prove that the limit of a function is infinity, we need to show that for any large positive number M, we can find a positive number δ such that whenever the input value z is within δ of i, the output value of the function will be larger than M. This can be done through algebraic manipulation and using the definition of a limit.
Yes, the limit can be proven using the epsilon-delta definition. This definition states that for any positive real number ε, we can find a positive real number δ such that whenever the input value z is within δ of i, the output value of the function will be within ε of infinity.
The significance of the function approaching infinity at z=i is that it means the function has a vertical asymptote at z=i. This indicates that the function has a singularity at z=i and the function values will become increasingly large as the input approaches i.
Yes, there are other ways to prove this limit. Some other common methods include using the Squeeze Theorem, the L'Hopital's Rule, or using the properties of limits such as the sum, difference, and product rules. It ultimately depends on the specific function and the tools available to the scientist.