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According to Wolfram Alpha:

http://www.wolframalpha.com/input/?i=1%2Ftan(pi%2F2)

It appears that this holds: $\frac{1}{\tan{(\frac{\pi}{2}})}=0$

We know that $\cot{\pi/2}=0$, but if written in the aforementioned form, does this really still hold?

I wouldn't think so, since as we are evaluating the expression, the denominator becomes undefined, and we can't just manipulate that into a 0 can we?

(I'm aware there are similar looking questions on this site, but they don't seem to address this specific problem)

Edit: Wolfram alpha also thinks 1/(1/0) = 0.

greenturtle3141
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  • You're on the right track $\pi/2$ is a singularity of $\frac{1}{\tan x}$ but it's removable because on a small enough neighborhood around $\pi/2$, the function is equal to $\cot x$. Thus $\lim_{x\to\pi/2}\frac{1}{\tan x}=0$ – N8tron May 27 '18 at 03:28
  • Yes, it is obvious the limit holds (as x approaches pi/2), but two things: 1. Yeah that's not an equality right? 2. Why does WA treat this as an equality? – greenturtle3141 May 27 '18 at 03:29
  • Look up the concept removable singularity. Basically the idea is functions can be formally extended to remove all of them. That's usually how we do business in calculus. But from a conceptual level you are correct, they aren't equal as functions. Wolfram is just taking the extension with the removable singularities removed – N8tron May 27 '18 at 03:33
  • I know what this is, but I'm asking this question on a basis of rigor. I'm writing mock precalculus final exams, and WA claiming stuff like 1(1/0) is defined is worrying me. If we were to consider a function such as $\frac{1}{\tan(x)}$, can we say that it is $\mathbb{R} - {\frac12x\pi \mid x\in\mathbb{Z}}$? – greenturtle3141 May 27 '18 at 03:39
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    Unless you're teaching from Wolfram alpha I think you're fine – N8tron May 27 '18 at 03:44
  • Why is $\frac{1}{\frac{1}{0}}$ undefined? – Vepir Apr 10 '22 at 20:42

2 Answers2

4

Wolfy means $$\lim_{x\to\pi/2}\frac1{\tan x}=0$$

Indeed, it is often ‘okay’ to accept $$\frac1{\pm\infty}=0$$

EDIT: $\frac1{\frac10}$ is undefined, but $\lim_{x\to0}\frac1{\frac1x}$ is defined.

For $x\ne0$, we have $\frac1{\frac1x}=x$.

Therefore, $$\lim_{x\to0}\frac1{\frac1x}=\lim_{x\to0}x=0$$

Szeto
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  • This is a true statement, but is this really the calculation as advertised? Like when I ask Wolfram Alpha to compute 1+1, they aren't going to find the limit as x approaches 1 of 1+x. Plus, when I show all steps of 1/(1/0), WA does some crazy thing with multiplying by 0/0. – greenturtle3141 May 27 '18 at 03:28
  • @greenturtle3141 Wolfy is made by human, and to make things simpler, they do not show steps rigorously. – Szeto May 27 '18 at 03:30
  • @greenturtle3141 Indeed, Wolfy is made for some advanced users and the idea of limit is often cosidered if the original expression is undefined and the limit is not shown explicitly. However, multiply by 0/0 is totally unacceptable. – Szeto May 27 '18 at 03:33
  • I can see this being acceptable once we're mathematicians and don't care about these tiny things. But in a classroom environment where rigor is extremely emphasized, would you consider a statement such as 1/(1/0) = 0 to be acceptable? – greenturtle3141 May 27 '18 at 03:41
  • @greenturtle3141 See my edit. – Szeto May 27 '18 at 04:09
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May be, you could look at the problem $$A=\frac{1}{\tan(x)}=\frac{\cos(x)}{\sin(x)}$$ Since $x$ approaches $\frac \pi 2$, let $x=\frac \pi 2+y$ which make $$A=-\frac{\sin(y)}{\cos(y}$$ Close to $y=0$, using equivalents $$A \approx -y$$ and then the result.

Claude Leibovici
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