Shape of Cotangent Function

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


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

Proof

 * $$\cot x$$ is continuous and strictly decreasing on $$\left({0 \, . \, . \, \pi}\right)$$:

Continuity follows from the Combination Theorem for Functions:


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

The fact of $$\cot x$$ being strictly decreasing on this interval has been demonstrated in the discussion on Cotangent Function is Periodic on Reals.


 * $$\cot x \to + \infty$$ as $$x \to 0^+$$:

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) $$\cos x \to 1$$ as $$x \to 0^+$$;
 * 2) $$\sin x \to 0$$ as $$x \to 0^+$$.

Thus it follows that $$\cot x = \frac {\cos x} {\sin x} \to + \infty$$ as $$x \to 0^+$$.


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

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

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

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


 * $$\cot x$$ is not defined and discontinuous at $$x = n \pi$$:

From the discussion of Sine and Cosine are Periodic on Reals, it was established that $$\forall n \in \mathbb{Z}: x = n \pi \Longrightarrow \sin x = 0$$.

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

Now, from the above, we have:
 * 1) $$\cot x \to + \infty$$ as $$x \to 0^+$$;
 * 2) $$\cot x \to - \infty$$ as $$x \to \pi^-$$.

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

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

From the periodic nature of $$\cot x$$, it follows that the same applies $$\forall n \in \mathbb{Z}: x = n \pi$$.

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


 * $$\cot \left({n + \frac 1 2}\right) \pi = 0$$:

Follows directly from Sine and Cosine are Periodic on Reals: $$\forall n \in \mathbb{Z}: \cos \left({n + \frac 1 2}\right) \pi = 0$$.