Is the Limit of x/x! as x Approaches 0 Equal to 0 or Does it Not Exist?

[Prove It]

An interesting question has been posted by Brilliant.org.

What is $\displaystyle \begin{align*} \lim_{x \to 0} \frac{x}{x!} \end{align*}$?My intuition tells me that the limit does not exist. My reasons for this are:

1. A limit can only exist if its left hand and right hand limits exist and are equal. Since the factorial function is only defined for nonnegative integers, and is thus not defined for any x < 0, how could it possibly have a left hand limit?

2. Again, the factorial function is only defined for nonnegative integers, and is thus discontinuous between 0 and 1. How could a right hand limit possibly exist when you can not approach 0 from the right along the function?

3. $\displaystyle \begin{align*} \frac{x}{x!} \equiv \frac{1}{\left( x - 1 \right) !} \end{align*}$ for all $\displaystyle \begin{align*} x \neq 0 \end{align*}$. As it is not possible to have $\displaystyle \begin{align*} (-1)! \end{align*}$ it would make sense that the limit can not exist.However, some people have been making the following (I believe flawed) arguments:

1. As the top and bottom are both defined at 0, and the denominator is nonzero, then the answer should be 0/0! = 0/1 = 0. I believe this argument is flawed because this rule where direct substitution is possible can only be used when the numerator and denominator are both CONTINUOUS around the limiting point. As the denominator is not continuous, I believe this should not be used.

2. That the factorial function is continuous in a neighbourhood around 0 if written as the gamma function. I believe this too is flawed because while the gamma function is equal to the factorial function at all nonnegative integers, the gamma function itself is NOT the factorial function! It's like saying that a thumb is a type of finger while a finger is not necessarily a type of thumb.

3. WolframAlpha seems to agree that the limit is 0.Would anyone like to chime in as to whether my intuition is correct, or if it is not, why the limit is 0?

 

[Euge]

Hi Prove It,

Usually, $x!$ represents the gamma $\Gamma(x + 1)$. Is this what the OP meant?

 

[Jameson]

I agree it seems to come down to how you define $x!$. If it's the product definition, then I would vote it doesn't exist. If by the Gamma function definition, then it does. You already noted that though, so really this is a question of "what is the correct definition of factorial?".

This is a topic I've recently pondered. I was talking to my students about how specific cases can be extended to more general cases, like how Newtonian mechanics was expanded under General Relativity (I believe, not a physics expert), or how the Pythagorean Theorem can be thought of as a special case of the Law of Cosines. The area of a right triangle, $A=.5bh$ is a special case of the more general $A=.5bh\sin(\theta)$. When something becomes expanded, should we go with the most encompassing definition or the initial one?

 

[I like Serena]

Hey Prove It,

It boils down to definitions.

The (ε,δ)-definition of the limit of a function stipulates a requirement for any $x$ with $0 < |x-0| < \delta$.
If we pick $0 <\delta < 1$, there is no such $x$, meaning we cannot find a limit $L$ that satisfies $|f(x) - L| < \epsilon$.
And if we pick $\delta = 1.5$, we have one such $x$, so we'd get the limit actually at $x=1$.

The (ε,δ)-definition also requires $0$ to be a limit point, meaning that every neighborhood that contains $0$ also has to contain another point in the domain that is distinct from $0$.
With the real line topology, this is not the case, so the limit would be undefined.
$$\lim_{x\to 0} \frac{x}{x!} = \text{undefined}$$
However, with the discrete topology, this is the case, and we get:
$$\lim_{x\to 0} \frac{x}{x!} = \frac{1}{1!} = 1$$

Btw, the (ε,δ)-definition of the limit of a function does not require both left and right limits to exist.
Since the left limit does not exist, the limit is equal to the right limit.Wolfram Alpha is not very careful with definitions, so we cannot rely on it.
In particular it makes everything a single-valued function of complex numbers, so $x!$ is implicitly interpreted as $\Gamma(x+1)$, which it is not.
That same behavior can also give some unexpected results when including for instance the $\log$ function.The factorial function is not defined for non-integer values.
However, if we choose to extend its definition to mean $\Gamma(x+1)$, we get yet another limit:
$$\lim_{x\to 0} \frac{x}{\Gamma(x+1)} = \frac{0}{\Gamma(0+1)} = 0$$To summarize, we need clarification which topology the domain has, and what we mean exactly by $x!$, before we can say what the limit is.
With the standard real line topology and the standard factorial, the limit is ill-defined.

 

[Deveno]

I partially agree with I like Serena, it does depend on one's definitions.

In particular, for a limit of $\dfrac{1}{x!}$ to exist, it has to be DEFINED in a neighborhood of $0$ (whether or not $0!$ is defined at all is immaterial, so the first argument posted in Prove It's post is clearly fallacious-to use that logic, $\dfrac{x}{x!}$ needs to be CONTINUOUS in a deleted neighborhood of $0$).

The thing is, it is possible to create any number of continuous functions $f$ such that $f(n) = n!$ for $n \in \Bbb Z^+$ (for example use a periodic function times a gamma function), and that does not tell us unambiguously how such a function may behave near $0$.

Although the answer also depends on the topology used; in this context, I feel it is safe to say the usual (Euclidean metric) topology of the real numbers is intended, here.

 

[I like Serena]

I partially agree with I like Serena, it does depend on one's definitions.

Heh. So what is it that you disagree with?
Note that I gave a summary that with the standard real line topology and the standard definition of the factorial, the limit is ill-defined.

 

[Prove It]

I agree it seems to come down to how you define $x!$. If it's the product definition, then I would vote it doesn't exist. If by the Gamma function definition, then it does. You already noted that though, so really this is a question of "what is the correct definition of factorial?".

This is a topic I've recently pondered. I was talking to my students about how specific cases can be extended to more general cases, like how Newtonian mechanics was expanded under General Relativity (I believe, not a physics expert), or how the Pythagorean Theorem can be thought of as a special case of the Law of Cosines. The area of a right triangle, $A=.5bh$ is a special case of the more general $A=.5bh\sin(\theta)$. When something becomes expanded, should we go with the most encompassing definition or the initial one?

To me it comes down to avoiding circular arguments.

For example, some people like to prove $\displaystyle \begin{align*} \lim_{x \to 0} \frac{\sin{(x)}}{x} = 1 \end{align*}$ by using the series representation of $\displaystyle \begin{align*} \sin{(x)} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots - \dots \end{align*}$. However, this would be circular reasoning as getting the MacLaurin Series for $\displaystyle \begin{align*} \sin{(x)} \end{align*}$ requires knowing the derivative of $\displaystyle \begin{align*} \sin{(x)} \end{align*}$, and to find the derivative one must evaluate $\displaystyle \begin{align*} \lim_{h \to 0} \frac{\sin{(h)}}{h} \end{align*}$, the exact same limit we are trying to find! To avoid the circular reasoning we must have a way to evaluate the limit that does not require derivatives or the use of the derivative (so no L'Hospital's Rule either). The Sandwich Theorem works nicely in this case.

So for your question about Pythagoras' Theorem being a special case of the Cosine Rule, that is actually circular reasoning as well, as how does one prove the Cosine Rule? By drawing in the height of the triangle that is perpendicular to the base of the triangle and applying Pythagoras!

View attachment 5571

So where can we avoid the circular reasoning? One must prove Pythagoras' Theorem in a different way, and then one can make the extension to the Cosine Rule.

View attachment 5572

So with your question about the area of a triangle, how does one get the rule $\displaystyle \begin{align*} A = \frac{1}{2}\,a\,b\sin{(C)} \end{align*}$? Is it not from drawing in the height of the non-right-angled triangle and evaluating the height using trigonometry? And does this not mean we already have to know that the area of a triangle is $\displaystyle \begin{align*} A = \frac{1}{2}\,b\,h \end{align*}$ (in fact, we are simply evaluating $\displaystyle \begin{align*} h = a\sin{(C)}\end{align*}$.

So how do we get away from the circular reasoning? We need to realize that the area of a parallelogram is equal to the area of the corresponding rectangle with the same height, and then realising that a triangle is simply half of a parallelogram. In my opinion we should NEVER start off general. We need to start specific, be able to show something without any doubt, and then extend it to a more general case.

This is also why I believe we should not simply assume that the factorial function is the gamma function. The factorial function was created because in probability and combinatorics, we often have to write down the product of a positive integer and all the positive integers before that. Then one wanted to know if it was possible to extend to a more continuous function.

 

[Deveno]

Heh. So what is it that you disagree with?
Note that I gave a summary that with the standard real line topology and the standard definition of the factorial, the limit is ill-defined.

When one sees the expression:

$\lim\limits_{x \to 0} \dfrac{x}{x!}$

for it even to make *sense* in the standard topology, one has to have $f(x) = x!$ defined on some (deleted) neighborhood $(-\delta,0) \cap (0,\delta)$ of $0$.

As I pointed out before, the problem is not that some definition could not be made, but that the possible definitions are by no means *unique*.

I think we both agree the limit is ill-defined, but for slightly different *reasons*. I *do* agree the factorial function is *not* the Gamma function (shifted by one), but I don't feel the Gamma function is even the "best" candidate-we need SOME candidate for an "extension" of the factorial function but NONE IS GIVEN in the problem as stated.

(I *do* note in passing that if one defines the factorial function ONLY on $\Bbb N$ (or $\Bbb Z^{+}$), the subspace topology does indeed give us a discrete topology, which makes your inclusion of it in your post relevant).