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#1 |
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![]() 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 trig-identities. 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 |
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#2 |
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![]() sorry, the is supposed to be infinity (positive infinity that is). Philchy |
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#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 ![]() |
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#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. |
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#5 |
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Fvck time for another beer I think.... |
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#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. |
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#7 |
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Aye . . .beer . . .brain hurts . . . Struzor Shadownite 53 Gnome SK |
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