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Prove the sequence $s_n$ = $\frac{3^n + (-2)^{n+1}}{3^{n-2}-2^{n-1}}$ converges and find its limit.

So far I've reduced the limit to the following:
$s_n = \frac{\lim_{n\to\infty}3^n - 2\lim_{n\to\infty}(-2)^n}{(1/9)\lim_{n\to\infty}3^n - (1/2)\lim_{n\to\infty}2^n}$

$\lim_{n\to\infty}(-2)^n$ doesn't converge since it alternates from positive to negative powers of 2, and the other 3 limits diverge to $\infty$.
I'm stuck where to go next and any help is appreciated.

I've been asked to prove this question using the $\epsilon - N$ definition.

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Hint:

Use equivalents:

$(-2)^{n+1}=o(3^n)$, hence $3^n+(-2)^{n+1}\sim_\infty 3^n$.

Similarly, $3^{n-2}-2^{n-1}\sim_\infty 3^{n-2}$, so that $$\frac{3^n+(-2)^{n+1}}{3^{n-2}-2^{n-1}}\sim_\infty\frac{3^n}{3^{n-2}}=9.$$

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  • $\begingroup$ I'm not familiar with equivalents, but are you just saying that since 3^n for large n is massive in comparison to 2^(n-1) and therefore the limit is equivalent to 3^n $\endgroup$ – pretty fly for a pi guy Oct 21 '16 at 0:50
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    $\begingroup$ More or less, but the last sentence is meaningless: you should say the limit is the same as the limit of $3^n$. Precisely, equivalence of $f(n)$ and $g(n)$ means the ratio $f(n)/g(n)\to 1$ as $n\to\infty$. Furthermore a function which has a limit, is equivalent to its limit, and conversely. Equivalence of functions is compatible with multiplication and division (not with addition or subtraction). $\endgroup$ – Bernard Oct 21 '16 at 0:58
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Consider $s_n$ = $\frac{2}{\frac{2}{9} - (\frac{2}{3})^n}$ + $\frac{(-1)^{n+1}}{\frac{1}{27}(\frac{3}{2})^{n+1} - \frac{1}{4}}$

What are the limits as n approaches $\infty$ of $(\frac{2}{3})^n$ and $\frac{(-1)^{n+1}}{(\frac{3}{2})^{n+1}}$?

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Divide top and bottom by the term that grows the fastest, which is $3^n$. Thus: $$ \frac{3^n + (-2)^{n+1}}{3^{n-2}-2^{n-1}} = \frac{3^{-n}}{3^{-n}}\frac{3^n + (-2)^{n+1}}{3^{n-2}-2^{n-1}} = \frac{1+ (-2)(-2/3)^{n}}{3^{-2}-(2/3)^{n}/2} $$ Since $(2/3)^n \to 0$ as $n \to \infty$, the limit equals $\frac{1}{3^{-2}} = 9$.

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