Limit of the serie, challenge?

avest

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Hello guys! I can't handle this problem:
n approaches infinity ((3n2 -2n +5)/(3n2 +n+2))^(-n+4)
wolframalpha link: http://www.wolframalpha.com/input/?...+((3n^(2)+-+2n++5))/((3n^(2)+++n++2)))^(-n+4)

Here is what i've done: after substitution we obtain 1 to the negative infinity so i change it to e^(-n+4)ln(((3n2 -2n +5)/(3n2 +n+2))) then i try to solve the limit of the exponent. i write it such that i have [0/0] to use de l'Hospital's rule. In the mean time i split the ln to the difference of logarithms to make differentiation easier and..nothing! Problem becomes more and more complex so i suppose i don't see somthing or i don't know some trick.
I would be very greatful if you helped me with this problem.

Best regards
 
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Hello guys! I can't handle this problem:
n approaches infinity ((3n2 -2n +5)/(3n2 +n+2))^(-n+4)
wolframalpha link: http://www.wolframalpha.com/input/?...+((3n^(2)+-+2n++5))/((3n^(2)+++n++2)))^(-n+4)

Here is what i've done: after substitution we obtain 1 to the negative infinity so i change it to e^(-n+4)ln(((3n2 -2n +5)/(3n2 +n+2))) then i try to solve the limit of the exponent. i write it such that i have [0/0] to use de l'Hospital's rule. In the mean time i split the ln to the difference of logarithms to make differentiation easier and..nothing! Problem becomes more and more complex so i suppose i don't see somthing or i don't know some trick.
I would be very greatful if you helped me with this problem.

Best regards
I might play around with a take on

\(\displaystyle e \,= \, \displaystyle{ \lim_{x \,\to \, \infty} }\) \(\displaystyle \left(1\, + \,\dfrac{1}{x}\right)^x \)

That is, write

3 n2 + n + 2 = 3 n2 - 2n + 5 + 3 (n - 3}

and

\(\displaystyle \dfrac{3 n^2 \,+\, n \,+ \,2}{3 n^2\, -\, 2n\, +\, 5} \,=\, 1 \,+\, \dfrac{1}{n\, t_n}\)

where

\(\displaystyle t_n \,=\, \dfrac{1 \,-\, \dfrac{2}{3n}\, + \,\dfrac{5}{3n^2}}{1\, -\, \dfrac{3}{n}}\)

and thus

\(\displaystyle \displaystyle{ \lim_{n\, \to\, \infty} }\) \(\displaystyle \dfrac{3 n^2\,+\, n\, +\, 2}{3 n^2 \,- \,2n \,+ \,5} \,=\,\) \(\displaystyle \displaystyle{ \lim_{n\, \to\, \infty} }\) \(\displaystyle a_n \left[\left(1\, + \dfrac{1}{n \, t_n}\right)^{n\, t_n}\right]^{1/t_n}\)

where

\(\displaystyle a_n\, =\, \left[\left(1\, +\, \dfrac{1}{n\, t_n}\right)^{-4}\right]\)
 
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I would let y = ((3n2 -2n +5)/(3n2 +n+2))^(-n+4)

then ln(y)= (-n+4)ln((3n2 -2n +5)/(3n2 +n+2))

now lim as n->oo (ln(y))=lim n->oo [(-n+4)ln((3n2 -2n +5)/(3n2 +n+2))]=(-oo)(0)

So rewrite to get 0/0 and then use l'hopital's rule on the right hand side but for simplicity first write ln((3n2 -2n +5)/(3n2 +n+2)) as ln(3n2 -2n +5) - ln(3n2 +n+2).

Eventually you arrive at lim n->oo (ln(y)) = 1.
What does this last line mean?
It means that as n approaches infinity that ln(y) approaches 1. But ln(y) approaches 1 is the same as y approaching e.
Putting this all together we arrive at lim n->oo (y) =e
 
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