# How to show that $\lim \limits_{x \to -\tfrac{\pi}{2}} \tan (x) = - \infty$

How can I show that $\lim \limits_{x \to -\tfrac{\pi}{2}} \tan (x) = - \infty$ and $\lim \limits_{x \to \tfrac{\pi}{2}} \tan (x) = \infty$?

I have to prove that $\tan$ is surjective from $-\tfrac{\pi}{2}$ to $\tfrac{\pi}{2}$ and I thought I could use the Intermediate value theorem.

Any suggestions would be appreciated.

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It depends on which side you approach from. $\lim_{x \to -\tfrac{\pi}{2}^{+}} \tan (x) = - \infty$ while $\lim_{x \to -\tfrac{\pi}{2}^{-}} \tan (x) = + \infty$. So $\tan(-1.57079)\approx -158057.9$ and $\tan(-1.5708)\approx 272241.8$ –  Henry Nov 26 '11 at 13:19

First note that $$\lim_{x \to \tfrac{\pi}{2}^-} \sin(x) = 1$$ and $$\lim_{x \to \tfrac{\pi}{2}^-} \cos(x) = 0.$$

therefore $$\lim_{x \to \tfrac{\pi}{2}^-} \frac{\sin(x)}{\cos(x)} = + \infty$$ since we have (by the definition of limit):

$$\forall \varepsilon_1, \exists \delta_1, \forall x < \tfrac{\pi}{2}, |x - \tfrac{\pi}{2}| < \delta_1 \to |\sin(x) - 1| < \varepsilon_1$$ and $$\forall \varepsilon_2, \exists \delta_2, \forall x < \tfrac{\pi}{2}, |x - \tfrac{\pi}{2}| < \delta_2 \to |\cos(x) - 0| < \varepsilon_2$$ which together imply

$$\forall y, \exists \delta_3, \forall x < \tfrac{\pi}{2}, |x - \tfrac{\pi}{2}| < \delta_3 \to \frac{\sin(x)}{\cos(x)} > y$$

by letting $\delta_3 = \delta_2(\tfrac{1}{y})$ here $\delta_2$ is considered as a function of $\varepsilon_2$ by skolemization.

verification: Given $\delta_3 = \delta_2(\tfrac{1}{y})$ we see that $|\cos(x)| < \frac{|\sin(x)|}{y} < \frac{1}{|y|}$ and since both functions $\sin$ and $\cos$ are positive for these values of $x$ we have $\frac{\sin(x)}{\cos(x)} > y.$

You can prove the $-\infty$ part similarly.

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Are you claiming that if $\lim_{x \to a^-} f(x) = 1$ and $\lim_{x \to a^-} g(x) = 0$, then $\lim_{x \to a^-} f(x)/g(x) = +\infty$? That ain't right. –  TonyK Nov 27 '11 at 17:22
@TonyK, that is proved here in the case of $a = \pi/2$, $f = \sin$ and $g = \cos$, do you have an objection to this proof? –  user16697 Nov 27 '11 at 17:47
Your "therefore" is premature. It looks like you are claiming that the third equation is a consequence of the first two. It looked like that to me, anyway. –  TonyK Nov 27 '11 at 18:19
@TonyK: Yes I state two hypothesis (which are direct consequences of continuity) and then claim "therefore X" then unfold all the definitions and show how to prove X given the hypothesis. –  user16697 Nov 27 '11 at 18:23

The simple facts are that $\lim \limits_{x \to -\frac{\pi}{2}} \tan(x)$ is not equal to $-\infty$, and $\lim \limits_{x \to \frac{\pi}{2}} \tan(x)$ is not equal to $+\infty$.

This is because in every neighbourhood of either $\pi/2$ or $-\pi/2$, $\tan(x)$ takes on arbitrarily large positive and negative values. So $\tan(x)$ has no limit at these points. If you want to specify that you are taking the limit as $x$ tends to $\pi/2$ from below, you have to make that explicit: $$\lim_{x \to \frac{\pi}{2}^-} \tan (x)$$ or $$\lim_{x \to -\frac{\pi}{2}^+} \tan (x)$$

That looks a bit ugly $-$ in fact, it's difficult to read $-$ but you can also just say, for example, "the limit of $\tan(x)$ as $x$ tends to $\pi/2$ from below".

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Since you're working only in the interval $(-\pi/2,\pi/2)$, the cosine function is always positive. The sine is positive in the right half of the interval and negative in the left half. You could think about the fact that the cosine approaches $0$ at the endpoints and the sine approaches $1$ at the right end and $-1$ at the left.
Alternatively (although this is not a rigorous proof without additional work) draw a right triangle in which the length of the "adjacent" side is $1$ and look at what happens to the "opposite" side as the angle goes from $-\pi/2$ to $\pi/2$, and remember that the "opposite" side would then be the tangent.