# Calculus + Factorial = ?

Recently, I've been messing around with factorials (mainly because I was just told that $$x! \approx (\frac{x}{e})^x\times\sqrt{2\pi x}$$). I came across a couple calculus problems involving the function. Since I am very new to calculus, I figured I should ask the community on how to get the answers and what they are, not just by typing them into Wolfram Alpha and seeing what comes out.

Since $$x!$$ increases so rapidly, I decided taking the $$x$$ root of $$x!$$. This gave me the function, $$\color{Red}{f(x)=x!^{\frac{1}{x}}}$$. That function is the red graph in the picture below.

I noticed that, even though $$f(0)$$ is undefined (since$$\frac{1}{0}$$ is undefined), it still appears to have a value. That is my first problem: to find

$\lim_{x\rightarrow 0} \color{Red}{f(x)}$

Next, I noticed that $$\color{red}{f(x)}$$ is not linear. However, it appeared to be linear, so I decided to graph the derivative, which gave me the function $$\color{Blue}{g(x)=\dfrac{\text{d}}{\text{d}x}~f(x)}$$. This is the blue graph in the picture above. Since $$\color{red}{f(x)}$$ approached linearity (?), I knew that $$\color{blue}{g(x)}$$ must have a limit as it approached infinity. That leads me to my second problem: to find

$\lim_{x\rightarrow \infty} \color{blue}{g(x)}$

The graph is at this link.

NOTE: The domains of both functions stop before 171 because 171! is a massive number, to the point that most online calculators can't handle.

Note by Blan Morrison
3 months, 1 week ago

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1. Stirling's formula is a good reference for the approximation. Note that this approximation holds for large values, and not necessarily for small values.

2. According to Wolfram Alpha, $$\lim_{x\rightarrow 0 } f(x)$$ is an interesting value.
It is not immediately apparant to me why this is true as yet.
Note: $$\gamma$$ is the Euler-Mascheroni constant.

3. Note that $$f(x) \approx \frac{x}{e}$$ for larget values. Hence, if there is any justice in the world (meaning that if the limit exists), it is most likely that $$\lim g(x) = \frac{1}[e}$$.

4. However, there is a slight flaw in your logic. Namely, the following statement is not true: "If $$| f(x) - g(x) | < \epsilon$$ for all $$x$$, then $$\lim g'(x) = \lim f'(x)$$.".
An extra condition will need to be added to arrive at "If $$f(x) \approx g(x)$$ (in some manner), then $$\lim g'(x) = \lim f'(x)$$."

Staff - 3 months, 1 week ago

I did some experimenting with factorials, here are some interested facts:

1. The limit as f(x) approaches 0 is equal to the negative root of the equation $$x!=2$$
2. The limit as g(x) approaches infinity is equal to this limit: $\large \lim_{x \to \infty}(x^{0.1}e^{-x})^{\frac{1}{x}}$ and also this limit: (x can be any postive real number) $\large \lim_{n \to 0}(x^{n}e^{-x})^{\frac{1}{x}}$

Those limits are derived from the gamma function, and the second one seem to evaluate to $$\frac{1}{e}$$

- 2 months, 3 weeks ago

$$n\not=1$$ too, obviously

- 2 months, 3 weeks ago

I updated the comment

- 2 months, 3 weeks ago