Shape of Tangent Function

Theorem
The nature of the tangent function on the set of real numbers $\R$ is as follows:


 * $\tan x$ is continuous and strictly increasing on the interval $\left({-\dfrac \pi 2 . . \dfrac \pi 2}\right)$
 * $\tan x \to + \infty$ as $x \to \dfrac \pi 2 ^-$
 * $\tan x \to - \infty$ as $x \to -\dfrac \pi 2 ^+$
 * $\tan x$ is not defined on $\forall n \in \Z: x = \left({n + \dfrac 1 2}\right) \pi$, at which points it is discontinuous
 * $\forall n \in \Z: \tan \left({n \pi}\right) = 0$.

Proof

 * $\tan x$ is continuous and strictly increasing on $\left({-\dfrac \pi 2 . . \dfrac \pi 2}\right)$:

Continuity follows from the Combination Theorem for Functions:


 * 1) Both $\sin x$ and $\cos x$ are continuous on $\left({-\dfrac \pi 2 . . \dfrac \pi 2}\right)$ from Nature of Sine Function and Nature of Cosine Function;
 * 2) $\cos x > 0$ on this interval.

The fact of $\tan x$ being strictly increasing on this interval has been demonstrated in the discussion on Tangent Function is Periodic on Reals.


 * $\tan x \to + \infty$ as $x \to \dfrac \pi 2 ^-$:

From Sine and Cosine are Periodic on Reals, we have that both $\sin x > 0$ and $\cos x > 0$ on $\left({0 . . \dfrac \pi 2}\right)$.

We have that:
 * 1) $\cos x \to 0$ as $x \to \dfrac \pi 2 ^-$
 * 2) $\sin x \to 1$ as $x \to \dfrac \pi 2 ^-$

Thus it follows that $\tan x = \dfrac {\sin x} {\cos x} \to + \infty$ as $x \to \dfrac \pi 2 ^-$.


 * $\tan x \to - \infty$ as $x \to -\dfrac \pi 2 ^+$:

From Sine and Cosine are Periodic on Reals, we have that $\sin x < 0$ and $\cos x > 0$ on $\left({-\dfrac \pi 2 . . 0}\right)$.

We have that:
 * 1) $\cos x \to 0$ as $x \to -\dfrac \pi 2 ^+$
 * 2) $\sin x \to -1$ as $x \to -\dfrac \pi 2 ^+$

Thus it follows that $\tan x = \dfrac {\sin x} {\cos x} \to - \infty$ as $x \to -\dfrac \pi 2 ^+$.


 * $\tan x$ is not defined and discontinuous at $x = \left({n + \dfrac 1 2}\right) \pi$:

From the discussion of Sine and Cosine are Periodic on Reals, it was established that $\forall n \in \Z: x = \left({n + \dfrac 1 2}\right) \pi \implies \cos x = 0$.

As division by zero is not defined, it follows that at these points $\tan x$ is not defined either.

Now, from the above, we have:
 * 1) $\tan x \to + \infty$ as $x \to \dfrac \pi 2 ^-$
 * 2) $\tan x \to - \infty$ as $x \to -\dfrac \pi 2 ^+$

As $\tan \left({x + \pi}\right) = \tan x$ from Tangent Function is Periodic on Reals, it follows that $\tan x \to - \infty$ as $x \to \dfrac \pi 2 ^+$.

Hence the left hand limit and right hand limit at $x = \dfrac \pi 2$ are not the same.

From the periodic nature of $\tan x$, it follows that the same applies $\forall n \in \Z: x = \left({n + \dfrac 1 2}\right) \pi$.

The fact of its discontinuity at these points follows from the definition of discontinuity.


 * $\tan \left({n \pi}\right) = 0$:

Follows directly from Sine and Cosine are Periodic on Reals: $\forall n \in \Z: \sin \left({n \pi}\right) = 0$.