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Mathematics 56 Online
OpenStudy (anonymous):

lim n -> infinity (2n+ 2^n) / (3n+3^n)

OpenStudy (anonymous):

l'hopital's rule.

OpenStudy (anonymous):

no cant use that.....

OpenStudy (anonymous):

is the answer 2/3?

OpenStudy (anonymous):

not sure if its 2/3 or zero?

OpenStudy (anonymous):

i doubt that. i think it should be zero. the first terms are irrelevant since they are linear and the rest is exponential. if we simply erase the first terms you get \[(\frac{2}{3})^n\]which certainly goes to zero. of course this is not a proof.

OpenStudy (anonymous):

The answer is 0I used Mathematica to get the answer. I dont know how to get there though

OpenStudy (anonymous):

yes the wolfram is also showing zero but I dont know how to get there

OpenStudy (anonymous):

hey should I use??? |a(n) - 0 | < epsilon ???

OpenStudy (anonymous):

i still don't know why l'hopital does not apply. take the derivative once, get \[\frac{2+2^nln(2)}{3+3^nln(3)}\]

OpenStudy (anonymous):

I know we can use it but its analysis course, so I have to find the limit without l'hopital rule

OpenStudy (anonymous):

do it again, get \[\frac{2^n (ln(2))^2}{3^n (ln(3))^2}\]

OpenStudy (anonymous):

ooh sorry.

OpenStudy (anonymous):

any ideas you have....??? but now I m sure its 0. so I m half way I think

OpenStudy (anonymous):

i give up :(

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