@ultralegend5385 I know this because tan(π/2) = 1/0, and since the left and right hand limits of 1/x as x approaches 0 are equal to -∞ and ∞, 1/0 can either be ∞ or -∞.

@RyanG If you input larger and larger negative numbers, atan(x) will be closer and closer to −π/2.

We can introduce a point $\infty$ and say $\tan(\frac\pi2)=\infty$. With the right notion of distance on $\mathbb R\cup\{\infty\}$, this extended version of $\tan$ is even continuous. And we can even extend $\tan$ to the complex plane and $\mathbb R\cup\{\infty\}$ to $\mathbb C\cup\{\infty\}$ to get a holomorphic function. So your idea is very reasonable. It's just not usually done in lower level courses.

@Vercassivelaunos Since tan(π/2) = 1/0, does this mean that 1/0 = ∞?

The precise way to say it would be that the continuous extension of $x\mapsto\frac1x$ as a function to the extended real line maps $0\mapsto\infty$. This continuous extension has so many good properties that it would in a way be reasonable to just say $\frac{1}{0}=\infty$, but the issue is that we are also often considering a different kind of extension of the real line, which also contains $-\infty$. And with this different extension, the map from above can't be extended continuously. Due to this, it's better to be very context sensitive when considering such extensions.

Cutting to the chase: I would refrain from saying $\frac{1}{0}=\infty$, unless the context of the extended real line is crystal clear.

@Vercassivelaunos So if we are working in the extended real number line, 1/0=∞ ?

Then that would be a reasonable definition to make, yes.

@Vercassivelaunos But if 1/0 = ∞ then 1 = 0 * ∞. Is this allowed in the extended real number line?

The extended number line is not a field, meaning that it doesn't play nicely with algebra. It does play nicely with geometry, though. And division on the number line shouldn't be interpreted as the inverse operation to multiplication, but as a geometric operation: the extended number line can be visualized as a circle, with 0 on the bottom, $\infty$ at the top, 1 to r right and -1 to the left. $\frac1x$ should be interpreted as mirroring each point at the axis going through 1 and -1. Not as the multiplicative inverse of $x$.

@Vercassivelaunos Do you know a reliable website where I can learn more about the extended real number line?

@Vercassivelaunos Also, are we talking about the projectively extended real number line or the affinely extended real number line?

Not really. But if you want to search for yourself, you should look for projective geometry. The extended real line is also known as the projective real line.

@Vercassivelaunos Actually, when I said 1/0 = ∞, I meant 1/0 = +∞.

@Vercassivelaunos Also, can we continue this discussion in chat?

Yes, we can do that.

Just to clarify, 1/0 is equal to +∞ if we work in the affinely extended real number line?

I'm not quite sure what you mean by affinely extended. Anyway, it's the projective extension I've been talking about. There is only one single point at infinity there, so no need to specify $+\infty$, since there is no $-\infty$ in this context.

Have a look at this article to find out what 'affinely extended' means: en.wikipedia.org/wiki/Extended_real_number_line

Ah, I see. Then no, in the affinely extended line, $\frac10=+\infty$ would be a bad choice, since this would make 1/x discontinuous. It only makes good sense in the projective extension.

Okay. Here is a proof of 1/0 = +∞ if we are working in the affinely extended line which I would like to share with you.

The integral between 0 and 1 of 1/x is divergent, so we could say that it is equal to +∞. If we use the method for calculating definite integrals, we should find out that the integral between 0 and 1 of 1/x is equal to In(1)-In(0), which is equivalent to 0-In(0). In the affinely extended line,

In(0) = -∞, so 0-In(0) = 0--∞=0+∞=+∞. Therefore, 1/0 = +∞.

What do you think about this proof?