@FlybyNight: You guessed right; this is indeed even more stupid than the other reference. They actually say that $\lim_ {x\to0^-}x^x=1$ (which argument they then sweep away), but in fact any negative real …

Jul 20, 2010 · Why is any number (other than zero) to the power of zero equal to one? Please include in your answer an explanation of why $0^0$ should be undefined.

(0 + x)x = 0x + xx (0 + x)x − xx = 0x 0x + xx − xx = 0x I want to get rid of both xx terms using axioms I know. The simplest for me is to use the distributive law x(a + b) = xa + xb along with the property that …

At the origin, the absolute value function "bends" - it goes from decreasing with a slope of -1 to increasing with a slope of 1.

You have basically two options. Allow complex arguments, then the limit doesn't exist, or restrict to positive [or non-negative] arguments. Then $\lim\limits_ {x\searrow 0} e^ {x\log x}$ settles the matter.

Here, you can see in the plot that the line approaches $1$ as it gets close to $x=0$, that's why this limit is equal to 1. And in both cases, $f (0) = \frac {0} {0}$.

@Arturo: I heartily disagree with your first sentence. Here's why: There's the binomial theorem (which you find too weak), and there's power series and polynomials (see also Gadi's answer). For all this, …

Nov 20, 2016 · I tried to let ϵ=0.5x, but find ϵ might be zero in this way, which is contradict to for all ϵ>0. Is this statement false?

Apr 1, 2020 · If we didn't consider constant polynomials to be polynomials, the standard description of polynomial division wouldn't work without a complicated circumlocution for what would become a …

Jan 19, 2021 · Hint You just have to complete the square. See that $$\frac {1} {2}a (t + \frac {v_0} {a})^2=\frac {1} {2}a (t^2 + 2\frac {tv_0} {a}+\frac {v_0^2} {a^2})=\frac {at^2 ...