Prove lim z->i ((z^2+i)/(z^4-1))=∞

Please, Give us something readable without us having to switch back & forth between tabs/windows.

 

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.

 

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.

[itex]\displaystyle \left(\frac{-1+i}{\sqrt{2}}\right)^2=-i[/itex] giving division by zero for the denominator of z² + i .

 

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.

 

I just skimmed through your PDF and I think I found a problem.

You established that [itex]|z^2 + i| < 11/2[/itex] but then you applied this with [itex]|z^2 + i|[/itex] in the denominator. In that case, you need to reverse the inequality:

[tex]\frac{1}{|z^2 + i|} > \frac{1}{11/2}[/tex]
Therefore your final inequality is not correct.

If you want an upper bound for [itex]1/|z^2 + i|[/itex], then you need a LOWER bound for [itex]|z^2 + i|[/itex].

 

By the way, here's what I was getting at earlier in message #3. Suppose you have proved that

[tex]\lim_{z \rightarrow i} \frac{z^4 - 1}{z^2 + i} = 0[/tex]
Then this means, by definition, that given N > 0 there exists [itex]\delta > 0[/itex] such that
[tex]\left| \frac{z^4 - 1}{z^2 + i} \right| < \frac{1}{N}[/tex]
whenever [itex]0 < |z - i| < \delta[/itex]. But that's exactly what you need, right? (Just take the reciprocal of the above inequality.) Therefore the problem reduces to showing that
[tex]\lim_{z \rightarrow i} \frac{z^4 - 1}{z^2 + i} = 0[/tex]
and you can do this quite simply by using the algebra of limits (i.e., the limit of the quotient is the quotient of the limits), instead of playing around with all these inequalities.

Of course, if your instructor intends for you to do a full delta-epsilon proof for everything, then you should carry on doing that. It's hard to tell from the problem statement.