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anonymous
 4 years ago
Show , lim x>0 ( sin 1/x) not equal to 0
use epsilon delta method
anonymous
 4 years ago
Show , lim x>0 ( sin 1/x) not equal to 0 use epsilon delta method

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it's lies in the interval 1,1 isn't ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0nothing like that , prove by epsilon delta , thats all i know ! :p

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2http://www.wolframalpha.com/input/?i=lim+x%3E0+%28+sin+1%2Fx%29 seems like undefined

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0is the ans there at this link ?

experimentX
 4 years ago
Best ResponseYou've already chosen the best response.2it says that limit is undefined

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but is it explained ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lol ! I also know the ans , but the prob is explanation ! :p

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This has a somewhat rigorous proof

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0This is a simple way to understand: http://mathforum.org/library/drmath/view/53369.html

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have many lyk these ! epsilon delta is really gettn on my nerves ! :p

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you are studying single variable or multivariate?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0both ! Actually in eco(h) , maths is subsi ! so its not core maths , but yea its maths , so bound to study ! :p

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lol, great :D :)) I wish I could help you more on epsilon delta proofs my current syllabus only requires to get the answer any how any ways ... so I tend to avoid rigorous stuffs.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lol, I have many many others things to study so not that :P

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hint if you take the sequence \[ x_n = \frac{2}{(2n+1)\pi}\] The sequence \[ x_n\] converges to zero but \[ \sin\left ( \frac 1{x_n} \right)=\pm 1\]
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