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factorial - Why does 0! = 1? - Mathematics Stack Exchange
$\begingroup$ The theorem that $\binom{n}{k} = \frac{n!}{k!(n-k)!}$ already assumes $0!$ is defined to be $1$. Otherwise this would be restricted to $0

Is $0$ a natural number? - Mathematics Stack Exchange
Inclusion of $0$ in the natural numbers is a definition for them that first occurred in the 19th century. The Peano Axioms for natural numbers take $0$ to be one though, so if you are working with these axioms (and a lot of natural number theory does) then you take $0$ to be a natural number.

What is the integral of 0? - Mathematics Stack Exchange
How about drawing sum upper and lower sums! You won't get very far because you'll be married to the horizontal axis and then, of course, all of the sums are zero and since a definite integral is always sandwiched between any upper and any lower sum. The value is trapped by 0. I.E. 0 <= the integral <= 0.

I have learned that 1/0 is infinity, why isn't it minus infinity?
1 x 0 = 0. Applying the above logic, 0 / 0 = 1. However, 2 x 0 = 0, so 0 / 0 must also be 2. In fact, it looks as though 0 / 0 could be any number! This obviously makes no sense - we say that 0 / 0 is "undefined" because there isn't really an answer. Likewise, 1 / 0 is not really infinity. Infinity isn't actually a number, it's more of a concept.

One divided by Infinity? - Mathematics Stack Exchange
The issue is implicit in the OP's phrasing and it is worth making it explicit. Namely, the OP is assuming that, just as $0.9$ or $0.99$ or $0.999$ denote terminating decimals with a finite number of 9s, so also $0.999\ldots$ denotes a terminating decimal with an infinite number of 9s, the said infinite number being denoted $\infty$. Changing ...

How do we calculate factorials for numbers with decimal places?
I was playing with my calculator when I tried $1.5!$. It came out to be $1.32934038817$. Now my question is that isn't factorial for natural numbers only? Like $2!$ is $2\\times1$, but how do we e...

limits - Prove that $\lim \limits_{n \to \infty} \frac{x^n}{n!} = 0 ...
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How to calculate 1 in _______ chance from a percentage?
So, from the first response, you divide 1 by the decimal if I am correct. For example: if I have 2%, I convert it into a decimal (0.02), then do 1/0.02, which equals 50 - giving the answer: 1 in 50 chance. $\endgroup$ –

Why is $\\frac{987654321}{123456789} = 8.0000000729?!$
So $\frac{8}{81}$ starts $0.098765\ldots$. The only break in the pattern is when you get to the digit '0' - subtracting 0 from 10 leaves you with 10, or a '1' in the next digit on the left, changing the 1 to a 2. So $$ \frac{8}{81} = 0.098765432098765432098765\ldots $$

Limit of $\\frac{x^c-c^x}{x^x-c^c}$ as $x \\rightarrow c$
My question is: Show that $\lim_{x \rightarrow c} \frac{x^c-c^x}{x^x-c^c}$ exists and find its value. Because the limit is 0/0 I've tried using L'Hopital's rule, but every time I differentiate it I

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