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Find: $\lim_{n\to{+}\infty}{{(2+n^3)}^{55-7n}}$ According to Maple, that is equal to zero. What theorem could I use?

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Put: $ y = {{(2+n^3)}^{55-7n}} $, then taking the natural log of both sides gives $ \ln (y) = (55-7n)\ln(2+n^3) $

Now, take the limit of both sides of the above equation:

$ \ln (\lim_{n \to \infty} y) = \lim_{n \to \infty} (55-7n)\ln(2+n^3) = - \infty $

Exponentiate both sides of the last equation:

$ \lim_{n\to\infty} y = 0 .$

Note that I used the following to find the limit: $ (55-7n)\ln(2+n^3) = \left( 55-7\,n \right) \left( 2\,\ln \left( n \right) +2\,{n}^{-2}- 2\,{n}^{-4}+O \left( {n}^{-6} \right) \right) $

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    Good answer +1.2012-07-16
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Note that

$\lim (2+n^3)^{55-7n}=\lim\frac{1}{ (2+n^3)^{7n-55}}$

Now, what is $\lim 2+n^3$? And what is $\lim 7n-55$?

That should get you started.

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    @mathsalomon Well, the easiest route I see now is the squeeze theorem, with 0<\frac{1}{ (2+n^3)^{7n-55}}<\frac{1}{n} for n>8. What I was trying to tell you before was how to think what the behaviour of the sequences was without being too rigorous.2012-07-16