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Sep
16
answered Solving one-sided limits analytically
Sep
10
comment Infinite amount of additions, finite sum?
@ZoltánSchmidt: While I found your question interesting, I think at some point you just need to accept that adding an infinite amount of terms can result in a finite result, and the intuition will come later. Someone on this site once quoted something along the lines of "In math, we don't understand things; we just get used to them". Try to work out the $\sum \frac1{2^n}$ example. It's easy to do and it will give you some insight as to why the total sum is $2$ and not $\infty$.
Sep
10
comment Is it true that $ \sum \limits_{i=1}^{\infty} f(i) = \lim_{n \to \infty} \sum \limits_{i=1}^{n} f(i) $?
By the way, it doesn't make sense to ask if this holds for all $n$, since $n$ isn't a free variable.
Sep
9
revised Why can't we substitute in limits for other limits?
added 2 characters in body
Sep
8
comment $f(x)=f(x^2+ 1/4)$ , $f$ is continuous from $\mathbb{R}$ to $\mathbb{R}$
$f$ can't be one-to-one, since for example $f(0) = f(\frac14)$.
Sep
7
answered Problem with differentiation as a concept.
Sep
7
comment Problem with differentiation as a concept.
Small correction: The definition is either $\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$ or $\lim_{x \to x_0} \frac{f(x)-f(x_0)}{x-x_0}$.
Sep
4
comment Integration of function
The binomial theorem proper only works when the power is a positive integer. You could get a series expansion, but otherwise I doubt this has a closed form. If it was a definite integral, you may be able to relate to the beta function using the substitution $u=\frac{a}{x}$.
Sep
4
comment Integration of function
Please make sure I did the LaTeX right. Is it $1+\frac{a}{x}$ or $\frac{1+a}{x}$?
Sep
4
revised Integration of function
latex
Sep
3
answered Reasoning Behind Holes in Rational Functions
Sep
2
comment Easy way to find the streamlines
If you're not supposed to do it rigorously, then sketching $\mathbf{F}$ at a few select points can give you an idea.
Sep
2
comment What's the Period of This Function?
Try graphing it, for particular values of $k$.
Sep
1
revised Limits/partial derivative
added 2 characters in body
Aug
30
comment The Inequalities
I'm not sure I understand the question, and I don't have the book. Would you mind rephrasing it a bit?
Aug
30
revised Finding the inverse of $y= 100(1-0.9^t)$
edited tags and LaTeX
Aug
24
comment why $x^2 = y^3$ is not smooth?
Usually you require that the derivative is never zero, which in this case happens at $t = 0$.
Aug
23
comment How to indicate keeping the sign when squaring a number.
While $x \cdot |x|$ certainly works, there's nothing wrong with the cases definition, and it's actually easier to understand.
Aug
16
comment Can $0\le\lim_{t\rightarrow+\infty}\int_{0}^tf(\tau)\mathrm{d}\tau\le a$ imply $\lim_{t\rightarrow+\infty}\int_{0}^tf(\tau)\mathrm{d}\tau$ exists?
The issue here is that you ask if from $0 \le \lim \int_0^t f \le a$ we can deduce the limit in question ($\lim \int_0^t f$) exists. But if it turns out it doesn't exist, then the question didn't make sense in the first place. I'm not sure what the best and rigorous way to describe the same meaning is, because the way I see it, there is no meaning. I hope this doesn't sound rude; I'm just trying to use the mathematical terms. The question as stated has no meaning because you are using something that you don't know exists in a way that can only be used if it in fact exists.
Aug
16
comment Can $0\le\lim_{t\rightarrow+\infty}\int_{0}^tf(\tau)\mathrm{d}\tau\le a$ imply $\lim_{t\rightarrow+\infty}\int_{0}^tf(\tau)\mathrm{d}\tau$ exists?
Since you accepted Clayton's answer, it would seem that $\int_0^t f(\tau)\ d\tau$ is indeed what you mean. The problem isn't the limit symbol; by definition, $\int_0^\infty f = \lim_{t\to \infty} \int_0^t f$. The problem is that by saying that this integral is between $0$ and $a$, you are implicitly assuming that it exists, otherwise you couldn't compare because you can't compare a number with something that doesn't exist. (cont.)