09202002, 12:14 PM  #1 
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A Calculus Question
With all the geniuses around here, I figured someone could help me do a extra credit proof for Calculus. Proove: Lim [x(sin(1/x))] = 1 x> Now for restrictions: 1) Brute Force Cannot Be Used 2) Verbal Proofs May Not Be Used Someone told me it could be done by some brutal manipulation with trigidentities. It's extra credit because most of the time in Calc, you assume 1/ = 0, but is obviously not the case. If you can come up with it and show it, what can I say. You'll have my deepest respect and thanks. Philchy 
09202002, 12:16 PM  #2 
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whoopsy
sorry, the is supposed to be infinity (positive infinity that is). Philchy 
09202002, 08:56 PM  #3 
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Lim [x(sin(1/x))] = 1 = 0* x> let's assume is infinity what you want is to get it 0/0 or / soo.... for 0* turn f*g into f/[1/g] or in this case [sin(1/x)]/[1/x] then if you take the limit as x>, you'll end up with 0/0 now using L'Hospital's Rule take the derivative of both f and g separately sooo...1/x = u du=[1/x^2]dx...d/dx sin u = cos u * du...you end up with...cos u du / du lim ([cos(1/x)][1/x^2]dx)/[1/x^2]dx x> obviously [1/x^2]dx cancel out and you are left with lim cos(1/x) x> u = 1/x lim 1/x = 0 x> and finally here we go... lim cos u = cos 0 = 1 u>0 oh yea!!!!! this is using Calculus B and i'm assuming your in Calculus A or something lower so u might not even know about derivatives yet 
09212002, 06:59 AM  #4 
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Note: Don't speak using "u" as "you" and "cos" as "because" if you're trying to explain a calculus problem. 
09212002, 07:41 AM  #5 
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Fvck time for another beer I think.... 
09212002, 07:07 PM  #6 
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Another way to do it is to expand the sin function in a Taylor's series. sin(u)= u  u^3/3! + u^5/5! ... so sin(1/x)= 1/x 1/(3! x^3) ... then x sin(1/x) = 1  1/(3! x^2) ... therefore lim [x sin(1/x)] = 1 x>inf The "correct" way to solve the problem really depends on whether you've learned l'Hopital's rule or Taylor's series in your class yet. 
09232002, 03:51 AM  #7 
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Aye . . .beer . . .brain hurts . . . Struzor Shadownite 53 Gnome SK 
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