Is $\displaystyle \lim_{x\to x_0} f(x)$ an operator? Is it legal to say; Let $\mathcal{L_{x_0}}$ denote $\displaystyle \lim_{x \to x_0$ then $L_{x_0} f(x)=\dots$?
@Balarka Since you're interested in algebraic geometry: Cohomology theories are how algebraic geometry is really done. The simplest, most pleasant early examples to learn first from come from algebraic topology. (And I might add that the topological covering space theory is really not going to be helping you much for etale covering spaces....)
In class my professor was talking about $(z^2+ \epsilon^2)^{\frac{1}{2}}$
And he said that since it was real valued on the real axis we could say that it had only real coefficients in it's expansion. Or something to that effect...
Well, you can define a branch of that on the complement of $[-\epsilon,\epsilon]$, for example. What's the point?
Oh, if you have an analytic function that's real on the real axis, you can compute all the derivatives at $0$ by using $\partial/\partial x$, and you will see that all the derivatives are real. That's true.
Is $\displaystyle \lim_{x\to x_0} f(x)$ a linear operator? Is it legal to say; Let $\mathcal{L_{x_0}}$ denote $\displaystyle \lim_{x \to x_0$ then $L_{x_0} f(x)=\dots$?
@Semiclassical OK. I ask as I received a better answer to my recently posted MSE question than any of the current ones (in person) in terms of Cantor normal form.
@Chris'ssis: I may have a way to extend this answer for $\sum\limits_{n=1}^\infty\frac{H^{(p)}_n}{n^q}$. The formula may not be very useful as it might be ugly when $p$ is bigger than $1$.
Don't know, that's all it gives. I have high suspicions it's a trick question, a competitor of mine irl told me she had a problem she couldn't solve and she would send it if I solved this one.
Doesn't my counter-example fulfil the requirements?
and the integral $\int_{-2}^2 10e^{-(x-0.5)^2} \mathrm{d}x$ is greater than $2$(in fact I don't know how to prove it elementary, but even if it isn't, the $10$ is a number I chose, it can be made arbitrarily large untill it is greater than $2$)
@nerdy Hint I take $R=\Bbb R$ the reals. It suffices you show that if a continuous function is not zero at $x$ there is a nbhd of $x$ where it is not zero.
@RandomVariable are you talking about $\displaystyle\lim_{d\to1^-}\log\left(\frac{1-d}{1-d^{c}}\right)$
The numerator is essentially $1$ the denominator is the thing that is changing enough to make a difference
@RandomVariable Think of it this way, the integral is between $$\lim_{d\to1^-}\int_d^{d^c}\frac{d^{b/c-1}}{1-t}\mathrm{d}t$$ and $$\lim_{d\to1^-}\int_d^{d^c}\frac{ d^{b-c}}{1-t}\mathrm{d}t$$ and the ratio of those is about $1$
@robjohn Yeah. You can't make a substitution just for the 2nd term unless you break up the integral. But you can't break it up if the upper limit is $1$.
@PedroTamaroff Let $A = \Bbb R[x,y]/\langle x^2+y^2-1\rangle$; $P = \langle x, 1-y\rangle$. I claim but will not prove that this is not isomorphic to $A$ (as $A$-mods) but that they're locally isomorphic.
