27
$\begingroup$

The way I have understood, $0/0$ is undefined or indeterminate because, if $c=0/0$ then $c\cdot 0=0$, where $c$ can be any finite number including $0$ itself.

If we also observe a fraction $F=a/b$ where $a,b$ are positive real numbers, the value of $F$ increases with the decrements of $b$. Being $0$ the least non-negative integer, if $b$ tends $0$ then $F$ tends to $\infty$ which is greater than all finite numbers.

I have also heard that no number is equal to infinity, a variable can tend to infinity. Any ratio is also a variable or is resolved to a number.

Now,

$$\tan(\pi/2) = \frac{\sin(\pi/2)}{\cos(\pi/2)}=\frac{1}{0},$$

is it undefined or infinity?

Also we know

$$\tan2x= \frac{2\tan x}{1-\tan^2x}$$

Is this formula valid for $x=\pi/2$?

Any rectification is more than welcome.

Some references:

$\endgroup$
10
  • 2
    $\begingroup$ It is undefined. $\endgroup$ Sep 1, 2012 at 11:56
  • 8
    $\begingroup$ A number does not tend to, a sequence or a function might. $\endgroup$
    – Stefan
    Sep 1, 2012 at 11:59
  • $\begingroup$ What is infinity? I don't remember infinity being defined. $\endgroup$
    – jimjim
    Sep 1, 2012 at 12:01
  • $\begingroup$ You are not computing limits, and therefore you cannot divide by zero. $\endgroup$
    – Siminore
    Sep 1, 2012 at 12:02
  • 1
    $\begingroup$ @Arjang, do you mean infinity and undefined are synonymous? $\endgroup$ Sep 1, 2012 at 12:03

3 Answers 3

28
$\begingroup$

In algebraic sense, it is just an undefined quantity. You have no way to assign a value to $\tan (\pi/2)$ so that its value is compatible with usual algebraic rules together with the trigonometric identities. Thus we have to give up either the algebraic rules or the definedness of $\tan(\pi/2)$. In many cases, we just leave it undefined so that our familiar algebras remain survived.

But in other situation, where some geometric aspects of the tangent function have to be considered, we can put those rules aside and define the value of $\tan(\pi/2)$. For example, where we are motivated by the continuity of the tangent function on $(-\pi/2, \pi/2)$, we may let $\tan(\pm\pi/2) = \pm\infty$ so that it is extended to a 1-1 well-behaving mapping between $[-\pi/2, \pi/2]$ and the extended real line $[-\infty, \infty]$. In another example, anyone who is familiar with complex analysis will find that the tangent function is a holomorphic function from the complex plane $\Bbb{C}$ to the Riemann sphere $\hat{\Bbb{C}}=\Bbb{C}\cup\{\infty\}$ by letting $\tan(\frac{1}{2}+n)\pi = \infty$ the point at infinity.

In conclusion, there is no general rule for assigning a value to $\tan(\pi/2)$, and if needed, it depends on which property you are considering.

$\endgroup$
0
22
$\begingroup$

$\tan \frac{\pi}{2}$ is undefined, $$\lim \limits_{x \to \frac{\pi}{2}^{+}} \tan x = -\infty \quad \text{and} \quad \lim\limits_{x \to \frac{\pi}{2}^{-}} \tan x = +\infty~,$$ where the ${}+{}$ and ${}-{}$ in the limits indicate whether a point is approached from right or left side.

In different contexts infinity may or may not be considered undefined but in general a thing being undefined does not imply it is infinite. In this case, note that $f(x_0)$ is entirely unrelated to $\lim \limits_{x \rightarrow x_0} f(x)$, unless the function is continuous at $x_0$.

$\endgroup$
4
  • $\begingroup$ Yes but isn't $tan(\theta)$= ratio of the length of the opposite side to the length of the adjacent side of a triangle. So even here some kind of limit is emerging! $\endgroup$ Sep 1, 2012 at 12:24
  • $\begingroup$ A limit is indeed emerging - namely the two I wrote down. A triangle with two $90^{\circ}$ angles can hardly be called a triangle, but even if you choose to do so, that only gives you $1 \over 0$. $\endgroup$ Sep 1, 2012 at 12:30
  • $\begingroup$ And if you do go for $\frac{1}{0}$ you may need consider $\frac{1}{\pm 0}$, i.e. $\pm \frac{1}{ 0}$ $\endgroup$
    – Henry
    Sep 1, 2012 at 13:03
  • $\begingroup$ So is it perfectly valid to write $\displaystyle\lim_{x\to \infty}\arctan {x}=\frac{\pi}{2}$ for the same reason? $\endgroup$ Jun 16, 2016 at 11:17
4
$\begingroup$

It depends on what you mean.

Is $\tan$ a real-valued (partial) function? Then $\tan(\pi/2)$ is undefiend.

Is $\tan$ an extended real-valued (partial) function? Then $\tan(\pi/2)$ is undefined.

Is $\tan$ a projective real-valued (partial) function? Then $\tan(\pi/2) = \infty$, and $\tan$ is actually a total function on the reals. (total meaning its domain is all of the reals)

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.