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## anonymous 5 years ago lim n -> infinity (2n+ 2^n) / (3n+3^n)

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1. anonymous

l'hopital's rule.

2. anonymous

no cant use that.....

3. anonymous

is the answer 2/3?

4. anonymous

not sure if its 2/3 or zero?

5. 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.

6. anonymous

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

7. anonymous

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

8. anonymous

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

9. anonymous

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

10. anonymous

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

11. anonymous

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

12. anonymous

ooh sorry.

13. anonymous

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

14. anonymous

i give up :(

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