Homework Help: Prove lim z->i ((z^2+i)/(z^4-1))=∞

1. Sep 18, 2012

thed0ctor

2. Sep 18, 2012

SammyS

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

3. Sep 18, 2012

jbunniii

Instead of trying to bound the expression above by 1/N, why not apply what you know about the algebra of limits?

$$\lim_{z \rightarrow i} \frac{(z + i)(z^2 - 1)(z - i)}{z^2 + i} = \frac{[\lim_{z \rightarrow i} (z + i)] [\lim_{z \rightarrow i} (z^2 - 1)][\lim_{z \rightarrow i} (z - i)]}{\lim_{z \rightarrow i} (z^2 + i)}$$

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.

4. Sep 18, 2012

SammyS

Staff Emeritus
thed0ctor,

See what bound $\displaystyle \left|z-i\right|<\frac{1}{2}$ puts on $\displaystyle \left|\frac{(z + i)(z^2 - 1)}{z^2 + i}\right|\ .$

5. Sep 18, 2012

thed0ctor

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.

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.

Attached Files:

• problem limit.pdf
File size:
1.1 MB
Views:
100
6. Sep 18, 2012

SammyS

Staff Emeritus
$\displaystyle \frac{-1+i}{\sqrt{2}}$ is less that 1 unit from i on the complex plane, so δ = 1 is too large.

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

7. Sep 18, 2012

thed0ctor

Sorry the last bit was meant to say

$$∴ \lim_{z \rightarrow i} \frac{z^2 + i}{z^4 -1} = ∞$$
not:
$$∴ \lim_{z \rightarrow i} \frac{z^2 + i}{z^4 -i} = ∞$$

Does this fix that issue? Because now I don't see the division by zero issue.

8. Sep 18, 2012

SammyS

Staff Emeritus
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.

9. Sep 18, 2012

thed0ctor

Good point. I also believe I found a bound. it's a nasty number but it's off the assumption that |z-i|< δ< .5

I attached the pdf. I think I'm getting better with inequalities. Thanks so much for your help thus far!

Attached Files:

• bounding.pdf
File size:
1.1 MB
Views:
70
10. Sep 19, 2012

jbunniii

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

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

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

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

11. Sep 19, 2012

jbunniii

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

$$\lim_{z \rightarrow i} \frac{z^4 - 1}{z^2 + i} = 0$$
Then this means, by definition, that given N > 0 there exists $\delta > 0$ such that
$$\left| \frac{z^4 - 1}{z^2 + i} \right| < \frac{1}{N}$$
whenever $0 < |z - i| < \delta$. But that's exactly what you need, right? (Just take the reciprocal of the above inequality.) Therefore the problem reduces to showing that
$$\lim_{z \rightarrow i} \frac{z^4 - 1}{z^2 + i} = 0$$
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.

Last edited: Sep 19, 2012