How can $\frac{1}{\ln(x)}$ be equal to $0$?

Question

High school calculus student checking in here – first time poster.

I got asked a question by one of my friends: If $\cot x = \frac{1}{\tan(x)}$, how could $\cot(x) = 0$? For this to be possible, he reasoned, $\tan(x)$ would have to be equal to infinity – and division by infinity does not work.

Rewriting $\cot(x)$ as $\frac{\cos(x)}{\sin(x)}$ makes the problem make sense, but doesn't explain why it doesn't make sense in the previous form.

Furthermore, as I was looking for more examples of similar things, I found that (at least according to desmos) $\frac{1}{\ln(x)}$ is satisfied by the coordinate $(0, 0)$. I cannot wrap my head around how this is possible, particularly since $\ln(x)$ is not even defined at $0$.

I wasn't able to find a good answer to this online, and we would really appreciate an elegant (i.e. understandable for high school students) explanation of why this is.

Answer 1

If you define

$$\cot(x):=\dfrac1{\tan(x)},$$ then the cotangent can never be $0$. But if you define it as

$$\cot(x):=\dfrac{\cos(x)}{\sin(x)},$$ then it is indeed zero for all $x$ such that $\cos(x)=0$. The second is the "best" definition, and

$$\cot(x)\equiv\dfrac1{\tan(x)}\equiv\dfrac{\cos(x)}{\sin(x)}$$ is not completely true.

Now if we look at

$$\frac1{\log(x)},$$

this expression is indeed undefined at $x=0$ because the logarithm is undefined. You can extend the definition in a way that makes the function continuous as follows:

$$\begin{cases}x=0\to\lim_{t\to0}\dfrac1{\log(t)},\\x>0\to\dfrac1{\log(x)}.\end{cases}$$

---

Note that a grapher cannot show you the difference between the definitions, because they differ in isolated points, which are "infinitely tiny".

Answer 2

$$x \mapsto \frac{1}{\ln(x)}$$ is indeed not defined for $x=0$, but the point is that it can be extended to a continuous function $f : [0,1) \rightarrow \mathbb{R}$ satisfying $f(0)=0$. More precisely, define $$f(x)=\left\{ \begin{array}{ll} \frac{1}{\ln(x)} & \mbox{if } 0<x<1 \\ 0 & \mbox{if } x=0 \end{array} \right. $$

You can prove that $f$, as defined, is a continuous function, and this is the only way to extend your original function to a continuous one at $x=0$.

Answer 3

Desmos draws the graph of a function at the points at which it is defined. Indeed, $\frac1{\ln x}$ is undefined at $0$; but, since $\lim_{x\to0^+}\frac1{\ln x}=0$, you see nothing peculiar at $(0,0)$. If you ask Desmos to draw the graph of, say $\frac{\sin(x)}x$ or $\frac{\cos(x)-1}{x^2}$, the same thing will occur.

And the equality $\cot x=\frac1{\tan x}$ is valid only when $\tan(x)\ne0$.

Answer 4

You have a removable singularity in your function definition using the division. While the function as written is not defined for some values, there is a single way how those gaps can be filled while making the function behave nicely around them.

Dealing with equations we tend to do such things all the time. When you simplify $\frac{a^2-b^2}{a-b}={a+b}$ wer tend to ignore the special case of $a+b$ where this is invalid. And we often can get away with it although it can come to bite us. In that sense we can say the two sides of the equation with the cot in it are the same, keeping in mind that in the case of two that's not literally true.