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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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2 Answers

up vote 3 down vote accepted

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. –  Iyengar Jul 16 '12 at 9:46
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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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$$\lim (2+n^3)=+\infty \quad \;;\;\quad \lim 7n-55=+\infty$$ Then, I get $$\frac{1}{{+\infty}^{+\infty}}.$$ But now, ¿how calculate ${+\infty}^{+\infty}.$? –  mathsalomon Jul 16 '12 at 6:23
    
@mathsalomon Well, you don't really get $\infty^\infty$. What that should be suggesting is that (after $7n>55$) you have to positive unbounded sequences, one which is raised to the power of another. This means that the limit of $P_n=a_n^{b_n}$ (the two sequences) will be $+\infty$ (meaning that given an arbitrarily large $M$ we can find an $N$ such that $P_n>M$). This means that we can make $1/P_n$ arbitrarily small, which means... –  Pedro Tamaroff Jul 16 '12 at 6:28
    
You mean I have to prove that $$a_n^{b_n}\to+\infty$$ using bounds? –  mathsalomon Jul 16 '12 at 6:42
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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. –  Pedro Tamaroff Jul 16 '12 at 6:47
    
thanks, I will try. –  mathsalomon Jul 16 '12 at 7:09
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