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Mathematics 18 Online
OpenStudy (anonymous):

using L'hopital's rule evaluate the limit as x approaches infinity (1-(3/x))^(4x)

OpenStudy (zarkon):

Start by writing \(\displaystyle\left(1-\frac{3}{x}\right)^{4x}\) as \(\displaystyle \exp\left(4x\cdot\ln\left(1-\frac{3}{x}\right)\right)\)

OpenStudy (anonymous):

ok i can get that far

OpenStudy (zarkon):

\[\displaystyle \exp\left(4x\cdot\ln\left(1-\frac{3}{x}\right)\right)= \exp\left(\frac{\ln\left(1-\frac{3}{x}\right)}{\frac{1}{4x}}\right)\]

OpenStudy (anonymous):

that's where i got stuck. i don't know what to do from there

OpenStudy (zarkon):

use l'Hospitals on the stuff on the inside of exp

OpenStudy (anonymous):

doesn't it equal 0/infinity?

OpenStudy (anonymous):

l'hospitals rule doesn't work for those circumstances right?

OpenStudy (zarkon):

0/0

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