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yes I believe we can choose epsilon, since it is negation..

thanks it makes sense to me..

sin(1/x) must be greater than epsilon, and it is true, how about 1/(2kpi)

@satellite73
hi, satellite73, can you check my comment here

yes..

it should be I guess \[|x-a|\implies |f(x)-L|\ge \epsilon\]

yes this is what I know..

yes ..

but it is gonna be contradiction right, and we are gonna say f(x) is not cont. at x=0..

can I use my example to get contradiction..sin(1/x)=0 but greater than 1/2

not greater than.

i should have said \(\forall \delta>0\)
\(\exists k\) such that \(\frac{1}{(2k+1)\pi}<\delta\)

it makes more sense now..

be careful here, you cannot write
\[\sin(\frac{1}{x})=0\] that makes no sense

thanks.. it must be grater than 0

yeah, I thought something like that.. how about 5)ii)