# 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 $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$
$\tan x \to +\infty$ as $x \to \dfrac \pi 2 ^-$
$\tan x \to -\infty$ as $x \to -\dfrac \pi 2 ^+$
$\tan x$ is not defined on $\forall n \in \Z: x = \paren {n + \dfrac 1 2} \pi$, 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 $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$:

Continuity follows from the Quotient Rule for Continuous Real Functions:

$(1): \quad$ Both $\sin x$ and $\cos x$ are continuous on $\openint {-\dfrac \pi 2} {\dfrac \pi 2}$ from Real Sine Function is Continuous and Cosine Function is Continuous
$(2): \quad$ $\cos x > 0$ 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 $\openint 0 {\dfrac \pi 2}$.

We have that:

$(1): \quad \cos x \to 0$ as $x \to \dfrac \pi 2^-$
$(2): \quad \sin x \to 1$ as $x \to \dfrac \pi 2^-$

From the Infinite Limit Theorem 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 $\openint {-\dfrac \pi 2} 0$.

We have that:

$(1): \quad \cos x \to 0$ as $x \to -\dfrac \pi 2 ^+$
$(2): \quad \sin x \to -1$ 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 $x = \paren {n + \dfrac 1 2} \pi$:

From the discussion of Sine and Cosine are Periodic on Reals, it was established that:

$\forall n \in \Z: x = \paren {n + \dfrac 1 2} \pi \implies \cos x = 0$

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:

$(1): \quad \tan x \to + \infty$ as $x \to \dfrac \pi 2^-$
$(2): \quad \tan x \to - \infty$ as $x \to -\dfrac \pi 2^+$

As $\map \tan {x + \pi} = \tan x$ 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 Tangent Function is Periodic on Reals, it follows that the same applies $\forall n \in \Z: x = \paren {n + \dfrac 1 2} \pi$.

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::

$\forall n \in \Z: \map \sin {n \pi} = 0$

$\blacksquare$

### Graph of Tangent Function 