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

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.

Where am I making mistake?

• In 1st approach you are taking a specific path. You have fixed $\theta$. So the approach is not really working into proving that it is one and the same limit you reach all along. Take $\theta=\frac{\pi}{2},0,\pi$. It is not all the same. Commented Oct 13, 2020 at 19:07
• What are the values of $-e^{-i2\theta}$ for $\theta =0$ and for $\theta=\pi/2$? Commented Oct 13, 2020 at 19:08
• Sir, $\theta$ is arbitrary. Then how it is specific? I have consider polar form of complex number in Approach I. Commented Oct 13, 2020 at 19:09
• @MarkViola Sir, Since I have considered polar form. Then for $\theta= 0$ and $\frac{\Pi}{2}$ the values are -1 , -i respectively. Commented Oct 13, 2020 at 19:13
• And now you see that the limit fails to exist. Commented Oct 13, 2020 at 19:15

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, $$f(x)\to_{x\to x_0}\infty \iff |f(x)|\to_{x\to x_0}+\infty$$

• Thank you very much, Sir. It greatly helped. Commented Oct 13, 2020 at 19:54

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 $$\infty$$.

• Thank you very much! Commented Oct 13, 2020 at 19:56