Javier
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 Sep 21 comment lim calculus problem with infinity Also, the nth-root of n goes to $1$. If you don't believe me, take the logarithm of the expression. Sep 21 comment lim calculus problem with infinity The fraction goes to $1$ but the exponent goes to $\infty$; this doesn't go to $1$ but rather is an indeterminate case, as evidenced by the well known fact that $(1+\frac1{n})^n \to e$. Sep 18 comment Does infinite time = time with no end = never? The short answer is that if it takes an infinite amount of time to reach point B, then the object never reaches it. Of course, in the real world, no matter how precise your measurement is, there will be some time after which the object is closer to B than your instrument can distinguish. 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 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 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 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 $k$. 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.) Aug 15 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? No, what you can say is that $f(t)$ is bounded; then, if for example you know that $f(t)$ is increasing, then you know that its limit exists and that $\lim\ f(t) \le \lim g(t)$. You can't talk about limits without first knowing if they exist. If (again) I'm understanding what you're saying correctly, your inequality should be $0 \le \int_0^t f(\tau)\ d\tau \le a$. Aug 15 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? Am I not understanding something? To compare the limit with other numbers, first it has to exist, or else the comparison makes no sense. Aug 15 comment Why does $\frac{\partial^2f}{\partial x \partial y} = \frac{\partial^2f}{\partial y \partial x}$ You need $f$ to be $C^2$, I believe. Aug 13 comment Are electrodynamic problems in the complex plane relevant to real life? Maybe you should ask this in Physics.SE? Aug 10 comment Is my proof correct about limit of $\sin\left(\frac{1}{x}\right)$? It's not a big deal, but you should mention what happens when $|A| > 1$.