F(z) limit using formal definition

  • Thread starter Thread starter Meggle
  • Start date Start date
  • Tags Tags
    Definition Limit
Meggle
Messages
16
Reaction score
0

Homework Statement


For each of the following choices of f(z) use the definition of a limit to obtain lim z-->0 f(z) or prove that the limit doesn't exist
(a) \frac{|z|^{2}}{z}

Homework Equations


Formal limit definition

The Attempt at a Solution


f(z) = \frac{|z|^{2}}{z}
f(z) = \frac{x^{2} + y^{2}}{x +iy}
So if z=(x,0), f(z)=\frac{x^{2} + 0}{x +i0} = x
Then f(z) --> 0 as (x,y) --> 0 along the real axis
And if z=(0,y), f(z)=\frac{0 + y^{2}}{0 +iy} = \frac{y}{i}
Then f(z) --> 0 as (x,y) --> 0 along the imaginary axis
So maybe lim z-->0 f(z)= 0
Suppose lim z-->0 f(z)= 0, then for each \epsilon >0 there exists \delta >0 such that 0 < |z - 0| < \delta implies |f(z) - 0| < \epsilon
...
Sooooo how do I figure delta out of that? I can't see how to simplify it or what to do next. It's due tomorrow, of course. :smile: And there's a (b) and a (c) , but think I could work them out if I could finish this one.

Also, can anyone tell me how to make the formulas update? I've changed all the SUP to curly brackets and carrots, but it won't seem to referesh. Edit never mind, seems they just don't show right on preview.
 
Last edited:
Physics news on Phys.org
Here's a suggestion:
\left| \frac{z}{|z|} \right| = 1
So if you write
|f(z)| = \left| |z| \cdot \left( \frac{z}{|z|} \right)^{-1} \right|
you can work in your delta.
 
Epsilon= delta. Try to justify it.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
Back
Top