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I can understand why $${\lim_{θ \to 90} \tan(θ) = ∞}$$

But since $$\tan90°=\frac{\sin90°}{\cos90°}$$

which results in $\frac{1}{0}$, I get confused, as $\frac{1}{0}$ could be both positive and negative infinity, which is one of the ways I think of its undefined nature. In such case, why do we consider the positive infinity scenario?

Thanks in advance. :)

forest
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    $\cos 90^\circ=0$, so the result of dividing by it is undefined in the real numbers – J. W. Tanner Apr 20 '20 at 01:12
  • $\lim\limits_{\theta\to{90^\circ}^-}\tan\theta=+\infty$, $\lim\limits_{\theta\to{90^\circ}^+}\tan\theta=-\infty$, – J. W. Tanner Apr 20 '20 at 01:13
  • It's not positive infinity, rather just "infinity". https://en.wikipedia.org/wiki/Projectively_extended_real_line – pw1822 Apr 20 '20 at 01:14
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    $\lim_{\theta\to 90^{\circ}}\tan(\theta)$ is not $+\infty$. If $\theta > 90^{\circ}$, $\cos(\theta) < 0$, so from the left the limit is $+\infty$, but from the right the limit is $-\infty$. This limit does not exist in possibly the most extreme way. – Cameron L. Williams Apr 20 '20 at 01:16
  • The use of $\infty$ in a limit equation is shorthand for saying the value gets arbitrarily large as the argument approaches the specified thing. Infinity is not a number and one needs to be careful discussing it in an arithmetic way. – j0equ1nn Apr 20 '20 at 02:06
  • As others note, you must be careful about left-hand and right-hand limits. Be that as it may ... If someone were to compose, say, a geometrically-flavored introduction to the trig functions (PDF link) that restricts discussion to first-quadrant angles only, then it would be perfectly reasonable to think of $\tan 90^\circ$ as $+\infty$ to help promote an intuitive understanding of the behavior of $\tan\theta$ in that restricted context. Of course, once other quadrants come into view, that thinking would need to be refined. – Blue Apr 20 '20 at 02:41

4 Answers4

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$\frac{1}{0}$ is undefined, and same is the case with $\tan 90^{\circ}$.

To show why, use the definition of division. It says $\frac{a}{b}=c$ when $a=bc$ holds and $a$ is unique when $b,c$ are kept fixed. But note that in case of $0$, you have $a \cdot 0= c$ holds whenever you set $c=0$ and $a\in \mathbb{N}$, a contradiction to uniqueness.

  • You might think of it with limits and stuffs as mentioned in comments, but i would like to be basic –  Apr 20 '20 at 01:16
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If $\theta$ is exactly $90^\circ$, then $\tan \theta$ is undefined because of division by zero.

If $\theta$ tends to $90^\circ$, namely $\theta -90^\circ$ is (non-zero) infinitesimal, then $\tan\theta=\infty$.

You said:

as $\frac{1}{0}$ could be both positive and negative infinity

This is correct only when you clearly understand its meaning. $\frac{1}{0}$ is not valid as an arithmetic expression, it is only valid as a symbol, denoting that the numerator tends to $1$ and the denominator is infinitesimal. Just as when we speak of a "$\frac{0}{0}$ indeterminate", we actually mean the limit status of an expression of the form $\frac{\text{infinitesimal}}{\text{infinitesimal}}$.

Y.Lin
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I put an answer here a couple of years ago that may help: Finding the Possible Values of y = tan(x).

The gist: as $\theta \rightarrow \dfrac {\pi}{2}$, the values grow to $\infty$ except for $\dfrac {\pi}{2}$.

$\tan \dfrac {\pi}{2}$ is undefined due to the definition of tangent, i.e. $\dfrac {\sin \frac {\pi}{2}}{\cos \frac {\pi}{2}} = \dfrac {1}{0}$ (and similarly for $\dfrac {3\pi}{2}$, except $\sin \dfrac {3 \pi}{2}$ is $-1$).

bjcolby15
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I know this is a really old question, but due to it being interesting and also because I like this approach and used to wonder too, I will provide an answer.

So, let's start with the definition you have, $$\tan \theta=\frac{\sin \theta}{\cos \theta}$$ Thus, we have $$\lim_{x \to 90°}{\frac{\sin x}{\cos x}}=\frac{\lim_{x \to 90°}{\sin x}}{\lim_{x \to 90°}{\cos x}}=\frac{1}{\lim_{x \to 0}{x}}$$ This obviously equals $$\lim_{x \to 0}{\frac{1}{x}}\mbox{, and as you intuitively said this equals limit shows}\cos x \mbox{ "equals" } \frac{1}{0}.$$



Continuing on, we prove $\frac{1}{0}$ is undefined.
First, we have: $$\lim_{x \to 0^+}{\frac{1}{x}=+\infty}\mbox{, while}\lim_{x \to 0^-}{\frac{1}{x}=-\infty}$$ A number cannot be approaching 2 numbers simultaneously, so we usually consider $\frac{1}{0}$, and consequently $\tan 90°$ as undefined. (I honestly don't get why you said it's $+\infty$ as even on the graph of $\tan x$ it approaches both $+\infty$ and $-\infty$) .

Another way of defining tangent is $$\tan x = \frac{\mbox{ opposite side }}{\mbox{ adjacent side }}\mbox{, which gives} \lim_{x \to \infty}{\tan x = \frac{1}{0}}\mbox{, and we can also proceed from there.}$$
Thanks!
P.S. I'm also relatively new to this topic, so please help edit my answer. Also, I just started learning $\LaTeX$ a week ago, so I might be really bad haha :)
Bulby2024
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