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i need to compute the limit of a sequence

$A_n= \frac{(n+2)^{2n}}{(n^2 - n - 6)^n} $

i'm stuck. i used the ratio test and got

$\frac{(n^2-9)^n(n+3)}{(n^2-4)^n(n-2)}$ then i'm stuck. help

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  • $\begingroup$ You should read the matlab manual, some tutorial or ask somewhere where matlab is discussed. Your question here is off-topic. $\endgroup$ Feb 9 '15 at 6:51
  • $\begingroup$ edited. i read the manual and followed everything but it's still wrong... D: $\endgroup$
    – C_I_
    Feb 9 '15 at 6:54
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Hint: Take $\log A_n$, and then use L'hospitale rule. Namely:

$\log A_n = \dfrac{\log (n^2+4n+4) - \log (n^2-n-6)}{\frac{1}{n}}$.

Can you take it from here ?

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  • $\begingroup$ is it $0$? i used 2 online limit calulators and got $e^5$ from both sites. i wrote a matlab code and it says $0$ (i suck at matlab). help $\endgroup$
    – C_I_
    Feb 9 '15 at 7:45
  • $\begingroup$ thank you so much! got it! but i'm curious, why is it not okay to divide by the largest power of n? $\endgroup$
    – C_I_
    Feb 9 '15 at 8:01
  • $\begingroup$ It won't help simply because you still have the exponent which is $n$. Note the dividing by the largest power of $n$ applies quite strictly to polynomials only and your expression is not polynomial because of the exponent of $n$. $\endgroup$
    – DeepSea
    Feb 9 '15 at 8:03
  • $\begingroup$ oh wow thanks. i've been getting this wrong since ever. can you please look at my other question? P.S. i missed the brackets in matlab $\endgroup$
    – C_I_
    Feb 9 '15 at 8:06
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$$a_n=\frac{(n+2)^{2n}}{(n^2-n-6)^n}=\\a_n=\frac{(n+2)^{2n}}{((n+2)(n-3))^n}=\\a_n=\frac{(n+2)^n(n+2)^n}{((n+2)(n-3))^n}=\\\frac{(n+2)^n}{(n-3)^n}=\\(\frac{n+2}{n-3})^n=\\(\frac{n-3+5}{n-3})^n=\\(1+\frac{5}{n-3})^n\\$$it is easier to use ratio test now!

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  • $\begingroup$ but the denominator is $n-3$... $\endgroup$
    – C_I_
    Feb 10 '15 at 15:29

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