haha

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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Originally posted by pkcRAISTLIN Yeah, I’d like to know what horrible, scarring incident in your childhood turned you into such an ignorant, intellectual-hating philistine?

ye, it's 3/5. I'll explain to you tomorrow at school.

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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quote:

Originally posted by RenderedDream it's cauchy's =P i think...

wtf, you're still alive and posting?

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Heh, calculus... Just sold my book to a younger student...

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quote:

Originally posted by RenderedDream it's cauchy's =P i think...

L'Hospital

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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?

2+2 is four

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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Originally posted by Coup Vlad wins. Flawless victory.

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

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?