lim n -> infinity (2n+ 2^n) / (3n+3^n)
l'hopital's rule.
no cant use that.....
is the answer 2/3?
not sure if its 2/3 or zero?
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.
The answer is 0I used Mathematica to get the answer. I dont know how to get there though
yes the wolfram is also showing zero but I dont know how to get there
hey should I use??? |a(n) - 0 | < epsilon ???
i still don't know why l'hopital does not apply. take the derivative once, get \[\frac{2+2^nln(2)}{3+3^nln(3)}\]
I know we can use it but its analysis course, so I have to find the limit without l'hopital rule
do it again, get \[\frac{2^n (ln(2))^2}{3^n (ln(3))^2}\]
ooh sorry.
any ideas you have....??? but now I m sure its 0. so I m half way I think
i give up :(
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