How does solving limits of a complex function differ from a normal function

Question

I am having quite the tough time solving limits of complex functions. I have tried solving both of these problems by taking the limits as $x$ and $y$ approach $0$ but have come up with the wrong answer both times and I am unsure what I am doing wrong.

$1)$ Prove that $$ \lim\limits_{z \to 0}\frac{\Re z}{|z|}$$ does not exist

I attempted to do this by rewriting the limit as $$ \lim\limits_{x,y \to 0}\frac{x}{\sqrt{x^2-y^2}}$$ and then take limits from each side. $$ \lim\limits_{x \to 0}\frac{x}{\sqrt{x^2-y^2}} = \frac{0}{\sqrt{-y^2}} = 0 $$ and $$ \lim\limits_{y \to 0}\frac{x}{\sqrt{x^2-y^2}} = \lim\limits_{y \to 0}{\frac{x}{\sqrt{x^2+0}}}=\lim\limits_{y \to 0}\frac{x}{x}=1 $$

Since the limits from both sides aren't equal, the limit must not exist. I feel, however, as if I am doing something fundamentally wrong with this yet I am unsure what it is.

$2)$ Prove that $$\lim\limits_{z \to 0}\frac{(\Re z)^2}{|z|}=0$$

I employed a similar method to what I used above here by taking the limits from both sides but, again, I feel I have done something wrong since my answer doesn't turn out correct.

I rewrote the limit as $$ \lim\limits_{x,y \to 0}\frac{x^2}{\sqrt{x^2-y^2}}$$

Taking limits from both sides gives $$ \lim\limits_{x \to 0}\frac{x}{\sqrt{x^2-y^2}}=\lim\limits_{x \to 0}\frac{x^2}{\sqrt{0^2-y^2}}=0$$ and $$\lim\limits_{y \to 0}\frac{x^2}{\sqrt{x^2-y^2}}=\lim\limits_{y \to 0}\frac{x^2}{\sqrt{x^2-0}}=\frac{x^2}{\sqrt{x^2}}=x$$

Clearly I have done the second question completely wrong as I should be getting both limits equal to zero, but I don't. If anyone could shed some light on where I have gone wrong or what method I need to use I would be very grateful.

Answer 1

You need to show that the limit, approached along any path, is the same. (And not just any straight-line path, e.g. along a coordinate axis! To show that a limit doesn't exist, it's enough to demonstrate that the limits are different when approached along two different paths, but the converse is false.)

Strictly, your first examples should be $\lim_{x \to 0} \lim_{y \to 0}$ and $\lim_{y \to 0} \lim_{x \to 0}$, rather than just $\lim_{y \to 0}$ and $\lim_{x \to 0}$, even though those limit expressions are constants so they do just evaluate to themselves.

The reason your second example is different is because you haven't taken $\lim_x \lim_y$, but only $\lim_y$, to obtain an expression in $x$. (Of course, what would it mean for the value of $\lim\limits_{z \to 0}\frac{(\Re z)^2}{|z|}=0$ to be $\Re z$, anyway? That's the same kind of syntax error as saying $\lim_{x \to 0}{x} = \frac{1}{2} x$.)

Answer 2

$|z| = \sqrt {x^2+ y^2}$ that aside proving $\lim_\limits{(x,y)\to (0,0)} \frac{x}{\sqrt {x^2 + y^2}} DNE$ will prove that $\lim_\limits{z\to 0} \frac{\Re z}{|z|} DNE$

On problem 2)

What you have is an okay start, but remember that $x$ will be going to $0$.

However, this technique will prove that limits do not exist, but will necessarily not prove that limits do exist. For the limit to exist it must have the same value along every possible path.

I would use something like $(\Re z)^2 \le |z|^2$ to put an upper bound on the numerator. Then show that $\lim_\limits{z\to 0} \frac {|z|^2}{|z|} = 0$