Shape of Tangent Function
Contents |
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 Quotient Rule for Continuous Functions:
- $(1): \quad$ Both $\sin x$ and $\cos x$ are continuous on $\left({-\dfrac \pi 2 \,.\,.\, \dfrac \pi 2}\right)$ from 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 $\left({0 \,.\,.\, \dfrac \pi 2}\right)$.
We have that:
- $(1): \quad \cos x \to 0$ as $x \to \dfrac \pi 2 ^-$
- $(2): \quad \sin x \to 1$ as $x \to \dfrac \pi 2 ^-$
- $\tan x = \dfrac {\sin x} {\cos x} \to + \infty$ 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:
- $(1): \quad \cos x \to 0$ 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 $x = \left({n + \dfrac 1 2}\right) \pi$:
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:
- $(1): \quad \tan x \to + \infty$ as $x \to \dfrac \pi 2 ^-$
- $(2): \quad \tan x \to - \infty$ 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: $\forall n \in \Z: \sin \left({n \pi}\right) = 0$.
$\blacksquare$
Also see
Sources
- K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 16.5 \ (2)$
