$\begingroup$ 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 $0 <k < n$. A reason that we do define $0!$ to be $1$ is so that we can cover those edge cases with the same formula, instead of having to treat them separately.

2013年3月15日 · 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.

$\begingroup$ This definition of the "0-norm" isn't very useful because (1) it doesn't satisfy the properties of a norm and (2) $0^{0}$ is conventionally defined to be 1. $\endgroup$ – Brian Borchers

2017年1月22日 · 1) x^a × x^b = x^a+b; for x = 0 and a = 0, you would get 0^0 × 0^b = 0^b = 0, so we can't tell anything -- except confirm that 0^0 = 1 still works here! 2) x^{-a}=1/{x^a} -- so when a = 0 , x^{-0} = 1/x^0 = x^0 , which again does work for 0^0 = 1 ; 3) {x^a}^b = x^{a×b} , thus x^(1/n) is the n-th root -- and 1/n = 0 for no value of n , so ...

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2015年9月11日 · On the contrary, those limits tell you that the limit of the entire quotient is $0$. This may be easier to see if you rewrite to $$ \lim_{x\to\infty} f(x)\frac1{h(x)} $$ where $\lim_{x\to\infty} f(x) = 0 $ and $\lim_{x\to\infty} \frac1{h(x)}=0 $, and the product of two functions that both have limit $0$ surely also has limit $0$.

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2019年11月27日 · The volume of the parallelepiped determined by the row vectors of the matrix is $0$. The system of homogenous linear equations represented by the matrix has a non-trivial solution. The determinant of the linear transformation determined by the matrix is $0$. The free coefficient in the characteristic polynomial of the matrix is $0$.

There is no general consensus as to whether $0$ is a natural number. So, some authors adopt different conventions to describe the set of naturals with zero or without zero. Without seeing your notes, my guess is that your professor usually does not consider $0$ to be a natural number, and $\mathbb{N}_0$ is shorthand for $\mathbb{N}\cup\{0\}$.

2018年3月13日 · 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