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lim x-27/x^(1/3)-3
x->27


and

lim x^(1/3)-1/x^(1/4)-1
x->1

i tried these and everytime i do it they come out to equal 0/0
which is wrong because our teacher told us it wasnt 0

2007-09-25 11:31:46 · 2 answers · asked by the_hockey_guy1990 1 in Science & Mathematics Mathematics

2 answers

L'hopital

What dose this mean, well my friend told me it means you take the derivitive of the top over the derivitive of the bottom. soooooooo

der of x-27 = 1

the der of x^(1/3)-3=1/3(x^-(2/3))

soooooo that means you have 1/1/3(x^-(2/3))

which means you have 3(x^(2/3)

there you go do the same thing for the bottom and you are set

2007-09-25 11:41:52 · answer #1 · answered by James L 2 · 0 0

Of course, at x=27 and x=1, the values are 0/0. The concept is that you have to find out which part of the fraction is approaching zero FASTER, the numerator or the denominator. I believe you have to take the derivitive of both parts of the fraction.
In the first, you would get 1/[1/3(x^(-2/3))] or
3 x^(2/3). The limit of this is 27.

2007-09-25 11:43:26 · answer #2 · answered by cattbarf 7 · 0 0

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