hidden mistake


by limitkiller
Tags: hidden, mistake
limitkiller
limitkiller is offline
#1
Feb6-11, 12:06 PM
P: 72
that is too sad
it have been 2 days and i coudnt find out where do i make mistake.
i wanted to prove:" (f(x)*g(x))'= g(x)*f(x)'+ g(x)'*f(x)". so:
(f(x)*g(x))'= lim h→0 ((f(x+h)*g(x+h)-f(x)*g(x))/h)

=lim h→0 ((f(x+h)*g(x+h))/h) - lim h→0 (f(x)*g(x))/h)

=[lim h→0 ((f(x+h))/h)*lim h→0 (g(x+h))] - [ lim h→0 ((f(x))/h) * lim h→0 (g(x))]

=[lim h→0 ((f(x+h))/h)* g(x)] - [ lim h→0 ((f(x))/h) * (g(x)]

= g(x)*[lim h→0 ((f(x+h))/h) - lim h→0 ((f(x))/h)]

= g(x)*[lim h→0 ((f(x+h)-f(x))/h)]

= g(x)*f(x)'
then (f(x)*g(x))'= g(x)*f(x)' !(?)!
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disregardthat
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#2
Feb6-11, 12:09 PM
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The limits lim h→0 ((f(x+h))/h) and lim h→0 ((f(x))/h) does not necessarily exist.
arildno
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#3
Feb6-11, 12:43 PM
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Hint:
[tex]0=f(x+h)g(x)-f(x+h)(gx)[/tex]

LeonhardEuler
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#4
Feb6-11, 01:05 PM
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hidden mistake


Just to expand on what other people have said, rules like "The limit of the difference is the difference of the limits" only apply when both limits exist. So it is not true that
[tex]\lim_{h\to 0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}= \lim_{h\to 0}\frac{f(x+h)g(x+x)}{h} -\lim_{h\to 0}\frac{f(x)g(x)}{h} = \infty - \infty[/tex]
(the last equality is assuming neither f nor g is 0 or has a 0 limit at x)
Likewise, splitting up limits like that only works when the limits each exist for addition, multiplication and division. The limit of the denominator also can't be 0 in the case of division.
limitkiller
limitkiller is offline
#5
Feb6-11, 03:04 PM
P: 72
thanks
that was so silly.


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