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$ \lim\limits_{n\to\infty}\frac{2^n}{3^{\frac{n}{2}}}$

I used wolfram to get the limit as follows: "lim n tends infinity 2^n/3^(n/2)".

And using L'Hospital's rule the result was: $\displaystyle\frac{2^{n+6}}{3^{\frac{n}{2}}}$, which tends to infinity.

My question is why does $\displaystyle\frac{2^{n+6}}{3^{\frac{n}{2}}}$ tend to infinity?

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    The L'Hospital part is wrong. The derivative of $2^x$ is $2^x \ln 2$, for example. (And by the way: If you didn't know how to take the limit of your original expression $2^n/3^{n/2}$, how do you know all of a sudden that $2^{n+6}/3^{n/2} \to \infty$? Computing that limit is essentially the same problem as the one you started with.)2011-01-20

2 Answers 2

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Hint: $\frac{2^n}{3^{n/2}} = \left(\frac{2}{\sqrt{3}}\right)^n$

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    @Wilbert: The key is the *distributive* law for exponents - see my answer.2011-01-20
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HINT $\rm \frac{2^n}{3^{n/2}}\ =\ \bigg(\frac{4}{3}\bigg)^{n/2}$

This is simply the exponents analog of pulling a common factor out of a sum:

$\rm a^n\ *\ b^{\:m\ \:n}\ \ = \ (a\ *\ b^{\:m}\:)^{\:n} $

$\rm a\ n + b\ m\ n\ = \ (a + b\ m)\ n $

So it's essentially a consequence of the distributive law for exponents.

The given problem is the special case $\rm\ \ \ a,b,m,n\ \to\ 1/3,\:2,\:2,\:n/2$

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    Thanks! I can't upvote yet because my reputation =(2011-01-20