1. ## Limits

Find the following limit if they exist, otherwise explain why they do not exist:

$\displaystyle \lim \frac{Re(z)}{z}$
with $\displaystyle z\rightarrow 0$

Since $\displaystyle Re(z) = x$

So $\displaystyle \lim \frac {x}{x+iy}$ ??
with $\displaystyle z\rightarrow 0$

How do i continue from here?

2. Originally Posted by Richmond
Find the following limit if they exist, otherwise explain why they do not exist:
$\displaystyle \lim \frac{Re(z)}{z}$ with $\displaystyle z\rightarrow 0$

Since $\displaystyle Re(z) = x$
So $\displaystyle \lim \frac {x}{x+iy}$ ??
with $\displaystyle z\rightarrow 0$
How do i continue from here?
Consider the limit along the path $\displaystyle x=y$. What is it?

Consider the limit along the path $\displaystyle y=0$. What is it?

What does that tell you?

3. Originally Posted by Plato
Consider the limit along the path $\displaystyle x=y$. What is it?

Consider the limit along the path $\displaystyle y=0$. What is it?

What does that tell you?
I'm not pretty sure but I would guess that the limit along the path x = y is 0 and the limit along the path y = 0 is 0 as well.

Thus the limit of this function does not exist?

4. Originally Posted by Richmond
I'm not pretty sure but I would guess that the limit along the path x = y is 0 and the limit along the path y = 0 is 0 as well.

Thus the limit of this function does not exist?
z-->0 implies (x,y)-->(0,0)

$\displaystyle \lim_{(x,y) \to (0,0)} f(x,x) = \lim_{(x,y) \to (0, 0)} \frac {x}{x+ix} = \frac {1}{1+i}$

$\displaystyle \lim_{(x,y) \to (0,0)} f(x,0) = \lim_{(x,y) \to (0, 0)} \frac{x}{x} = 1$

Since different paths to the origin give different limits, the limit does NOT exist