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

anjali_pant
 2 years ago
Best ResponseYou've already chosen the best response.1nothing like that , prove by epsilon delta , thats all i know ! :p

experimentX
 2 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

anjali_pant
 2 years ago
Best ResponseYou've already chosen the best response.1is the ans there at this link ?

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

anjali_pant
 2 years ago
Best ResponseYou've already chosen the best response.1but is it explained ?

anjali_pant
 2 years ago
Best ResponseYou've already chosen the best response.1lol ! I also know the ans , but the prob is explanation ! :p

FoolForMath
 2 years ago
Best ResponseYou've already chosen the best response.1This has a somewhat rigorous proof

FoolForMath
 2 years ago
Best ResponseYou've already chosen the best response.1This is a simple way to understand:http://mathforum.org/library/drmath/view/53369.html

anjali_pant
 2 years ago
Best ResponseYou've already chosen the best response.1I have many lyk these ! epsilon delta is really gettn on my nerves ! :p

FoolForMath
 2 years ago
Best ResponseYou've already chosen the best response.1you are studying single variable or multivariate?

anjali_pant
 2 years ago
Best ResponseYou've already chosen the best response.1both ! Actually in eco(h) , maths is subsi ! so its not core maths , but yea its maths , so bound to study ! :p

FoolForMath
 2 years ago
Best ResponseYou've already chosen the best response.1lol, 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.

FoolForMath
 2 years ago
Best ResponseYou've already chosen the best response.1lol, I have many many others things to study so not that :P

eliassaab
 2 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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