Confusion in calculating $\lim\limits_{z\to0}\frac{-1}{z^2}$

Question

I tried to calculate the limit of $f(z) = \frac{-1}{ z^{2} } $ near zero using two approaches in the complex space.

First Approach (Using Polar form of Complex Number)

Take $z = re^{i\theta },$ where $\theta$ is arbitrary, to consider every possible path.

Then we get $$ f(z) = \frac{-1}{ z^{2} } = \frac{ -e^{i(-2 \theta )} }{ r^{2} } $$

So, $\lim\limits_{ z\rightarrow 0} \frac{-1}{ z^{2} } = \lim\limits_{ r\rightarrow 0} \frac{ -e^{i(-2 \theta )} }{ r^{2} } = -\infty$

Second Approach

Take $z = x + iy,$ then we get $$f(z) = \frac{-1}{ z^{2} } = \frac{-1}{ x^{2} - y^{2} + i(2xy) }$$

Now, we have to consider $$ \lim\limits_{x \rightarrow 0 ,y \rightarrow 0 } \frac{-1}{ x^{2} - y^{2} + i(2xy) }$$

i) When I choose a path $y \rightarrow 0$, followed by $x \rightarrow 0$ , so I got $$\lim\limits_{x \rightarrow 0} = \frac{-1}{ x^{2} } = -\infty $$

ii) When I choose a path $x \rightarrow 0$, followed by $y \rightarrow 0$, so I got $$\lim_{y \rightarrow 0} = \frac{-1}{ -y^{2} } = \frac{1}{ y^{2} } = \infty $$

Hence, since we obtain two different limits, along two different path, the limit doesn't exist.
(Contradictory to Approach I ).

Where am I making mistake?

Answer 1

You should know that:

1. When you calculate the limit of the function $f:X\rightarrow\mathbb{C}$, where $X$ is a metric space you consider $\mathbb{C}$ as a subset of $\overline{\mathbb{C}}=\mathbb{C}\cup\{\infty\}$ which is commonly known as Riemann sphere. You add just one point, namely $\infty$, which you should visualize as a point which surrounds the plane.

2. Assume you calculate the limit $\lim_{z\to z_0}f(z)$ of the function $f:A\rightarrow X$, where $A$ is a subset of $\mathbb{C}$ (or $\mathbb{R}^2$ because they have the same topology) and $X$ is a metric space. The limit is then actually equivalent to $\lim_{(x,y)\to (x_0,y_0)}f(x,y)$, where $z_0=x_0+iy_0$. But existence and equality of limits $$\lim_{x\to x_0}\left(\lim_{y\to y_0}f(x,y)\right), \lim_{y\to y_0}\left(\lim_{x\to x_0}f(x,y)\right)$$ does not imply the existence of the actual limit.

That being said, in your second approach, values of the function lie in $\mathbb{C}$ so $$\lim_{x\to 0}\left(\lim_{y\to 0}\frac{-1}{x^2-y^2+2ixy}\right)=\infty$$ and $$\lim_{y\to 0}\left(\lim_{x\to 0}\frac{-1}{x^2-y^2+2ixy}\right)=\infty$$ and it is the same $\infty$, namely the point which "surrounds" the plane. It does not however imply that the limit exists.

In your first approach values of the function are also in $\mathbb{C}$ so the answer can't be $-\infty$. Also changing to polar coordinates is tricky. Your can't just say that $$\lim_{z\to 0}\frac{-1}{z^2}=\lim_{r\to 0}g(r)$$ for some function $g$. Besides in the expression $$\lim_{r\to 0}\frac{-e^{i(-2\theta)}}{r^2}$$ we have variable $\theta$ on which the result may presumably depend.

EDIT. To answer the question about existence of this particular limit, we have to refer to the definition of topology on $\overline{\mathbb{C}}$. It is enough to know that the family of sets $\mathcal{B}_{\infty}=\{A_M:M >0\}$, where $$A_M=\{z\in\mathbb{C}:|z|>M\}\cup\{\infty\}$$ is the (topological) basis at point $\infty$.

It is true that for a function $f:X\rightarrow\mathbb{C}$, where $X$ is a metric space, $$f(x)\to_{x\to x_0}\infty \iff |f(x)|\to_{x\to x_0}+\infty$$

Answer 2

The limit of complex functions is defined in terms of modulus $|\cdot|$. Thus a function $f(z)$ is said to have a limit of $\infty$ at $z_0$ if for every arbitrarily large real $M$ we can find a poitive number $\epsilon(M)$ such that $|z-z_0|\lt\epsilon(M)$ implies that $|f(z)|\gt M$. Using this definition, given some arbitrarily large $M$ we can chose $\epsilon (M)<\sqrt M$ such that for all $z$ in the complex plane that obey $|z-0|\lt \epsilon (M)$ we have $|f(z)|=|\frac{1}{z^2}|>M$.

This definition comes from the projection of the complex plane onto the Riemann Sphere where all points that "infinitley away" from the center $z=0$ are projected onto the "north pole" of the sphere which corresponds to $\infty$.