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math help (pg. 2)
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| colonelcrisp |
now bear in mind i havent taken calc for over 4 years but from what i remember, using l'hopital's rule, (take derrivative of top and bottom)
your limit approaching zero is aprox 3/5 (three fifths) |
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| TheDarkOne |
| ye, it's 3/5. I'll explain to you tomorrow at school. |
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| RenderedDream |
| quote: | Originally posted by colonelcrisp
now bear in mind i havent taken calc for over 4 years but from what i remember, using l'hopital's rule, (take derrivative of top and bottom)
your limit approaching zero is aprox 3/5 (three fifths) |
it's cauchy's =P
i think... |
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| paranoik0 |
| quote: | Originally posted by RenderedDream
it's cauchy's =P
i think... |
wtf, you're still alive and posting? |
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| Cloudburst |
| Heh, calculus... :D Just sold my book to a younger student... |
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| OrZonE |
| quote: | Originally posted by RenderedDream
it's cauchy's =P
i think... |
L'Hospital |
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| _Nut_ |
| quote: | Originally posted by UWM
3x/(2x^2+5x) = 3x/2x^2 + 3x/5x = 3/x + 3/5
lim x-->0 (3/2x + 3/5)
Ding. |
Your epiphany makes perfect sense now. Or was the the Taylor series derivation you had earlier? |
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| BrownTA4Life |
| 2+2 is four:crazy: |
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| Vlad |
| quote: | Originally posted by colonelcrisp
now bear in mind i havent taken calc for over 4 years but from what i remember, using l'hopital's rule, (take derrivative of top and bottom)
your limit approaching zero is aprox 3/5 (three fifths) |
If my memory serves me right...
dx = n(x^n-1)º + n(x^n-1)¹
º = Outside
¹ = Inside |
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| sandstorm03 |
d (tan(3x))
/dt = 3/(cos3x)^2
d(2x^2 +5x)
/dt = 4x + 5
3/(cas(3x)^2)+4x+5) x-> 0 cos(3x)^2+4x->0
3/5 |
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| Yan |
| quote: | Originally posted by sandstorm03
d (tan(3x))
/dt = 3/(cos3x)^2
d(2x^2 +5x)
/dt = 4x + 5
3/(cas(3x)^2)+4x+5) x-> 0 cos(3x)^2+4x->0
3/5 |
Errr... Wha? :p |
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