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cinar Group TitleBest ResponseYou've already chosen the best response.0
@helder_edwin
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
first one i can do negate the definition by picking \(\epsilon =\frac{1}{2}\) then show that there does not exist a \(\delta\) such that \(x<\delta\implies \sin(\frac{1}{x})<\frac{1}{2}\)
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
that is because no matter how small \(\delta\) is there will be some \(k\) such that \[\frac{2}{(2k+1)\pi}<\delta\] and for that \(k\) and all other larger you will have \(\sin(\frac{1}{x})=1\)
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
yes I believe we can choose epsilon, since it is negation..
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
thanks it makes sense to me..
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
sin(1/x) must be greater than epsilon, and it is true, how about 1/(2kpi)<delta we have sin(1/x)=0 not greater than epsilon. can we use this..
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
@satellite73 hi, satellite73, can you check my comment here
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
hi do you know exactly how to negate the definition of \[\lim_{x\to a}f(x)=L ?\]that is what you need to use
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
definition of \[\lim_{x\to a}f(x)=L \] \(\forall \epsilon> 0\), \(\exists \delta \) such that \(xa\implies f(x)L<\epsilon\)
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
it should be I guess \[xa\implies f(x)L\ge \epsilon\]
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
negation is \(\exists \epsilon >0\) such that \(\forall \delta\), \(\exists x\) such that \(xa<\delta\) and \(f(x)L>\epsilon\)
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
yes this is what I know..
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
you need to use what i wrote above (took me a while) so the logic is this you exhibit a specific \(\epsilon\) and then show that no matter how small \(\delta\) is, then you can find some \(x\) for which \(xa<\delta\) and at the same time for that \(x\) you have \(f(x)L>\epsilon\) my suggestion was to pick \(\epsilon =\frac{1}{2}\) whatever you pick, it should be specific not general
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
in your example, you have to show that \(\exists \epsilon>0\) so that \(\forall \delta\) \(\exists x\) with \(x<\delta\) and \(\sin(\frac{1}{x})>\epsilon\)
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
but it is gonna be contradiction right, and we are gonna say f(x) is not cont. at x=0..
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
since \(\sin(\frac{1}{x})=1\) infinitely often close to zero, you can show this pick \(\epsilon=\frac{1}{2}\) for example, then \(\forall \delta>0\) \(\exists \frac{1}{(2k+1)\pi}<\delta\) and so \(\sin(\frac{(2k+1)\pi}{2})=1>\frac{1}{2}\)
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
you are negating the definition, but this is not a proof by contradiction, it is a direct proof, using the negation of the definition
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
a proof by contradiction would be to assume it is continuous and then arrive at some contradiction, but you do not need to do this here. write down exactly what the negation of the definition is, then prove it directly
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
can I use my example to get contradiction..sin(1/x)=0 but greater than 1/2
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
not greater than.
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
i should have said \(\forall \delta>0\) \(\exists k\) such that \(\frac{1}{(2k+1)\pi}<\delta\)
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
it makes more sense now..
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
be careful here, you cannot write \[\sin(\frac{1}{x})=0\] that makes no sense
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
again, i stress work the directly from the definition of " \(f\) is not continuous at \(a\)" i know it seems like a proof by contradiction, but it is not the definition of "\(f\) is not continuous at \(a\)" i wrote above
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
thanks.. it must be grater than 0
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
question 2 looks annoying but i think you can do it by saying something like this for any two points you can join them by a straight line (you can even write down what the line would be) unless the line would contain those two points, in which case you can join them by a horizontal and vertical line that skips those points probably an easier way, not sure
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
yeah, I thought something like that.. how about 5)ii)
 2 years ago

cinar Group TitleBest ResponseYou've already chosen the best response.0
for 5) ii) can we say that since A is compact it is finite.. I know it is bounded and closed.. is it true that every closed and bounded set is finite..
 2 years ago
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