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 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

sorry, the �‡ is supposed to be infinity (positive infinity that is).

Philchy

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

Note: Don't speak using "u" as "you" and "cos" as "because" if you're trying to explain a calculus problem.

Fvck time for another beer I think....

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.

Aye . . .beer . . .brain hurts . . .




Struzor Shadownite 53 Gnome SK