Limit when $z$ goes to $0$ of $re(z) / im(z)$?

Question

It's clear to me that this limit does not exist, because you can go to zero using the identity line and you get $1$, but if you go to zero in direction getting close to the imaginary line then you get $0$.

How can i formalize that idea, especially going to zero from the identity line?

Answer 1

You could write it as $$ \lim_{r\rightarrow 0} \frac{\mathrm{Re}(re^{i\theta})}{\mathrm{Im}(re^{i\theta})} $$ in which $\theta$ specifies the direction you approach 0.

This particular limit is equal to $\cot \theta$, and is indeed equal to 1 on the identity line $\theta = \pi/4,-3\pi/4$.

Answer 2

In Complex Analysis it's good to think visually, and then write the algebraic details:

We will do a proof by contradiction. Suppose there is a number $L$ somewhere on the complex plane with the following property: For any $\epsilon >0$, there exists a $\delta > 0$ such that for every $z \in D(0, \delta)$ we have $f(z) \in D(L, \epsilon)$. Now we are going to pick $\epsilon = \dfrac 13$ and try to show that given this $\epsilon$, there is no satisfactory $\delta$. Now, why did we pick $\dfrac 13$? Because you got $0$ and $1$ as your limits; no matter where $L$ is, you cannot have both $0$ and $1$ in a disk of radius $\dfrac 13$ around $L$.

Can you finish this proof and translate it into algebra, or do you need more details?

Answer 3

"go to zero using" is a bit flaky and informal, but you have the general idea.

The definition of $\lim_{z\to a}f(z) = K$ means that for any $\epsilon > 0; \epsilon \in \mathbb R$ you can find a $\delta > 0$ so that for all $z$ so that $|z-a| < \delta$ it will have to follow that $|f(z) - K| < \epsilon$.

Now the negation of the that is: For any of $\delta > 0$ there will always be $z_0$ where $|z_0-a| < \delta$ but $|f(z_0) - K| \ge \epsilon$ for some $\epsilon > 0$.

So "go to zero using the identity line" means: If $\delta > 0$ we will always be able to find some let $z_0 = a + ai$ where $0 < a < \frac {\delta}{\sqrt 2}$.

In this case, $|z_0 - 0| = |z_0| = \sqrt{2a^2}=\sqrt 2 a < \sqrt 2\frac \delta {\sqrt 2} = \delta$. For this $z_0$ we have $\frac {Re(z_0)}{Im(z_0)} = \frac aa=1$.

so if $\lim_{z\to 0} \frac {R(z)}{Im(z)} = K \ne 1$ we can set $\epsilon = |1 - K|> 0$ and then even though $|z_0 - 0| < \delta$ we have $|\frac {Re(z_0)}{Im(z_0)} - K| = |1-K| = \epsilon \ge \epsilon$.

That is a contradiction. so IF $\lim_{z\to 0} \frac {R(z)}{Im(z)}$ exists, it must be that $\lim_{z\to 0} \frac {Re(z)}{Im(z)} = 1$.

But "go to zero using the imaginary line" means: If $\delta > 0$ we can always find a $z_1 = 0 + ai$ where $0 < a < \delta$. Thus we have $|z_1 - 0| = |ai| = a < \delta$. But $\frac {Re(z_1)}{Im(z_1)} = \frac 0a = 0$. So $|\frac {Re(z_1)}{Im(z_1)} - 0| = |0| = 0$.

so if $\lim_{z\to 0} \frac {R(z)}{Im(z)} = K \ne 0$ we can set $\epsilon = |K| > 0$ and then even though $|z_0 - 0| < \delta$ we have $|\frac {Re(z_0)}{Im(z_0)} - K| = |0-K| =|K| = \epsilon \ge \epsilon$.

That is a contradiction. so IF $\lim_{z\to 0} \frac {R(z)}{Im(z)}$ exists, it must be that $\lim_{z\to 0} \frac {Re(z)}{Im(z)} = 0$.

So we have that IF $\lim_{z\to 0} \frac {Re(z)}{Im(z)}$exists, then it must be that it is equal to both $1$ and $0$. This is impossible.

So $\lim_{z\to 0} \frac {Re(z)}{Im(z)}$ can not exist.