Is the reciprocal of tan(pi/2) equal to zero, or is Wolfram Alpha insane?

Question

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.

Answer 1

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

Answer 2

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.