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1d
reviewed Close Complex Fourier coefficients and series
1d
reviewed Leave Open Probability Distribution Function?
1d
reviewed Close Number of chains having k subsets
Oct
18
reviewed Leave Open Distribution of things on one row.
Oct
18
reviewed Close Elementary Conditional Probability Question
Oct
18
reviewed Close How to wirte this sentence into Predicate
Oct
18
reviewed Close How to prove $l^p$ is separable?
Oct
18
reviewed Leave Open proof that inf(-A)=-sup(A)
Oct
18
reviewed Close graph of connected set
Oct
15
answered For small $x$, one has $\ln(1+x)=x$?
Oct
15
awarded  Good Answer
Oct
13
answered Is this induction proof correct?
Oct
12
awarded  Nice Answer
Oct
11
comment Calculating limit in parts. Why possible?
@CareBear: I guess my point is that the two sides of the equality are equal. It doesn't necessarily matter how you get that they are equal. When you have a rule saying that you can "do something" all it means is that two expressions are equal. So in this case you can "evaluate" by taking the limit of $\frac{f(x) - f(1)}{x-1}$ as $x$ approaches $1$. My argument, of course, doesn't extend to other cases. The comment above for the other answer suggests that one, in fact, doesn't have a general rule for this.
Oct
11
comment How to prove for every $ε > 0$, it holds that $0 \le a < ε$, then $a = 0$
Do you know how to use the @ symbol when writing comments? If you comment on an answer without using the @ symbol, then the answerer will assume that the comment is meant for him/her.
Oct
11
answered Calculating limit in parts. Why possible?
Oct
11
comment How to prove for every $ε > 0$, it holds that $0 \le a < ε$, then $a = 0$
@Pelto: You just rewrote my proof... Note that assuming that $a>0$ is the same as assuming that $a\neq0$. My proof is a proof by contradiction.
Oct
11
comment How to prove for every $ε > 0$, it holds that $0 \le a < ε$, then $a = 0$
@Did: Sorry, I didn't see this before answering ...
Oct
11
comment How to prove for every $ε > 0$, it holds that $0 \le a < ε$, then $a = 0$
@FarahFai: Exactly, so the assumption that $a>0$ must be wrong, and $a$ must be $0$.
Oct
11
answered How to prove for every $ε > 0$, it holds that $0 \le a < ε$, then $a = 0$