Shape of Tangent Function

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


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

Proof

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

Continuity follows from the Combination Theorem for Functions:


 * 1) Both $$\sin x$$ and $$\cos x$$ are continuous on $$\left({-\frac \pi 2 \, . \, . \, \frac \pi 2}\right)$$ 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 $$x \to \frac \pi 2 ^-$$:

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

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

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


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

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

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

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


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

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

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 $$x \to \frac \pi 2 ^-$$;
 * 2) $$\tan x \to - \infty$$ as $$x \to -\frac \pi 2 ^+$$.

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

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

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

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: $$\forall n \in \mathbb{Z}: \sin \left({n \pi}\right) = 0$$.