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I understand that this is a very stupid question but I'm not getting the answer.

At $x=\pi/2$, what is the value of $tan(x)$? Should it be $-\infty$ or $+\infty$?

Text tells it to be $+\infty$. But why?

Geometrically thinking, it comes out to be $+\infty$. But how to explain the graph which has both the values at $\pi/2$?

Widawensen
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kryptoknight
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  • $\displaystyle{\large\lim_{x \to \left(\pi/2\right)^{\pm}}\tan\left(x\right) = \mp\infty}$. – Felix Marin Oct 22 '13 at 22:20
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    For functions like $\tan$, it is not appropriate to extend the real line by two infinities, a positive and a negative one. One extends it by a single point at infinity, bending the line into a circle. Then you have a well-defined value $\tan \dfrac{\pi}{2} = \infty$. – Daniel Fischer Oct 22 '13 at 22:30
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    Your texts are wrong. You should burn them. $\tan(\pi/2)$ and $\tan(-\pi/2)$ are undefined (unless further context is given). If you want to define $\tan(\pi/2)$ like Daniel Fischer did, using the some form of the "extended real numbers", you have to state that this is what you're doing, and know what you're doing. – Stefan Smith Oct 22 '13 at 23:11
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    http://math.stackexchange.com/questions/189621/is-tan-pi-2-undefined-or-infinity – lab bhattacharjee Oct 23 '13 at 07:02

4 Answers4

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It shouldn't be anything because it's not defined there.

Clive Newstead
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Instead of trying to "evaluate" $\tan(\pi/2)$ by simply observing its graph, why not go back to good old-fashioned trig arguments? If you were to evaluate the value of $\tan(\pi/2)$ on the unit circle, you would essentially be constructing this argument: $$ \tan(\pi/2) = \frac{\sin(\pi/2)}{\cos(\pi/2)} = \frac{1}{0} $$ Note that is the same as asking for the $y$-coordinate of the point (0,1) on the unit circle to be divided by its $x$-coordinate. Due to various reasons (see http://en.wikipedia.org/wiki/Division_by_zero; or Numberphile's wonderfully informative video Problems with Zero http://www.youtube.com/watch?v=BRRolKTlF6Q), mathematicians have not defined what it means to divide by zero because the value is not consistent in various contexts. So division by zero is simply "undefined" - which leads to your uncertainty as to whether $\tan(\pi/2) = +\infty$ or if $\tan(\pi/2)=-\infty$ (!)

Xoque55
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There are two common ways to "compactify" the reals.

The first is the one you're probably more familiar with. You add two additional points, $+\infty$ and $-\infty$, located at the two "ends" of the number line. This is called the extended real line. Commonly, the notation $\infty$ is used instead of $+\infty$: thast causes some confusion with....

The second is called the projective real line. It adds only single point $\infty$, which is located on both ends of the line (think of the two ends being glued together to form a circle). This is closely related to the Riemann sphere and "complex infinity". This construction tends to be more useful when you're working with polynomials and rational functions, or when doing complex analysis. If we view $\tan$ as function from the usual real line to the projective real line, then we can continuously extend it so that $\tan(\pi/2) = \infty$.

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It depends on how you interpret the situation.

Case 1:-

We see $\displaystyle\tan\frac{\pi}{2}=\frac{\sin\frac{\pi}{2}}{\cos\frac{\pi}{2}}=\frac{1}{0}$ which is undefined in mathematics.

Case 2:-

As you can see from the graph the $\displaystyle\lim_{x\rightarrow\frac{\pi}{2}^{\mp}}\tan(x)=+\infty$.

Both cases are equally correct and valid.

Easily speaking it is geometrically infinite and algebraically undefined.

Soham
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    It is not indeterminate. It is undefined. And the limit must be unique, to exist: one needs to stick with one-sided limits. – Vincenzo Oliva Jul 19 '16 at 16:21
  • @VincenzoOliva If $\frac{0}{0}$ is indeterminate why shouldn't $\frac{1}{0}$ be? – Soham Jul 19 '16 at 16:29
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    Remaining in the context of limits: the first form can give any result, the second only an infinite one - we just can't tell the sign, so it remains undefined if we're not in the projective real line (see Hurkyl's answer above). In particular, the terms indeterminate and undefined are not interchangeable: $0^0$ is indeterminate but almost always defined as $1$. – Vincenzo Oliva Jul 19 '16 at 16:35