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anonymous
 5 years ago
Lim(x>0)[1/x]^sinx
anonymous
 5 years ago
Lim(x>0)[1/x]^sinx

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that form is \[\infty\]^0, so u have to use exp function

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0The answer given is 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Why don't you do a series expansion in x and then send the function to zero?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Actually, I should have worked it out before opening my mouth.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0exp[lim(x>0) sin x . 1/x] exp[lim(x>0) (ln x)/sin x] and then do l'hospital

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Actually, what I said will work, but you'd have to prove other limits first.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i do thinking about the concept before typing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0What are you thinking, iamignorant?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, I have got the idea that suzi gave. With that idea I am coming at e^(0x1)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0So that is the answer that I needed. Actually I did everything he did except applying lHospital

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is there a rank higher than HERO? =D like a grandmaster or something

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Who knows? This site's operations are a mystery half the time.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Incidentally, I set the RHS equal to y then took the log of both sides, put the result into an indeterminate form, used L'Hopital's rule and took the limit to zero after I suggested expanding the function. Just sayin'...
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