Recent content by pwsnafu

Here's a direct method ##\lim_{x\to-\infty}\left(1+\frac{1}{x}\right)^x = \lim_{y\to\infty}\left(1-\frac{1}{y}\right)^{-y} = \lim_{y\to\infty}\frac{1}{\left(1-\frac{1}{y}\right)^y}= \lim_{y\to\infty}\left(\frac{y}{y-1}\right)^{y}.## Now let ##w = y-1##. Clearly ##y\to\infty## is equivalent to...

Of course not, infinity is not a real number. However you can rigorously prove: Let ##f(x)## diverge to infinity and ##g(x)## diverge to negative infinity as ##x \to \infty##. Then ##(fg)(x)## diverges to negative infinity. If you have done epsilon proofs, then you should be able to do this.

No, it comes from solving equations. Suppose we are given ##a \geq 0## and we wish to find ##x \in \mathbb{R}## such that ##x^2 = a##. Then we show: If ##a = 0## there only exists one solution, namely ##x=0##, If ##a \neq 0## then there exists two distinct solutions ##x_1## and ##x_2##, We...

If that's your definition of fundamental, then addition of natural numbers is not fundamental. We define addition as ##a+S(b) = S(a+b)## where ##S## is the successor operator.

Also, I want to point out is not true. The left and right limits are not the same.

Yeah, sorry. I realized and edited my post. You replied just before I finished.

1/x is not defined at x=0. This means that ##f(a)## does not exist, hence you can't appeal to continuity (continuity requires a function to be defined there).

No, ##\aleph_1## is the cardinality of all countable infinite ordinals. The cardinality of real numbers is ##2^{\aleph_0}##. These are equal only if the continuum hypothesis is true.

Clarification: a "sequence" is a function whose domain is the natural numbers. PeroK is using "function" in place of the phrase "a function whose domain and co-domain are the real numbers".

And what do you do about the non-computable real numbers?

In Peano, addition is defined through the successor operator. So, addition is not fundamental either.

No it shouldn't. There should be no +C because fractional calculus is a theory of definite integrals.

Consider the ring of all operators ##C^\infty \to C^\infty##. What is the inverse of the derivative?

The Dirac measure is defined so that if a subset ##A \subset \mathbb{R}## contains the point 0 then it has measure 1. Otherwise the measure is 0. I'm going to write this as ##\delta(A)##. Now the Lebesgue integral in our case is (without going into details) ##\int_A f \, d\delta =...

Oh, the limit you want to evalute is ##\left(\frac{n+1}{n-1}\right)^{n}## and not ##\frac{n+1}{n-1}##. Yes that's ##1^\infty## indeterminate form. This is what happens when students do it naively: ##\lim_{n\to\infty}\left(\frac{n+1}{n-1}\right)^{n} =...