# limit of $2^n/3^{n/2}$ as $n\to\infty$

$$\lim_{n\to\infty}\frac{2^n}{3^{n/2}}$$

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: $\displaystyle\frac{2^{n+6}}{3^{n/2}}$, which tends to infinity.

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

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@Wilbert: Please use mark-up. – Arturo Magidin Jan 20 '11 at 4:02
What is mark-up? – Wilbert Barrera Jan 20 '11 at 4:03
@Wilbert: LaTeX, like you used in your first line, to make things clearer instead of relying on words like "as n goes to infinity". – Arturo Magidin Jan 20 '11 at 4:04
Oh that was the code I used in wolfram. That's why I enclose it in a code block. – Wilbert Barrera Jan 20 '11 at 4:08
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.) – Hans Lundmark Jan 20 '11 at 7:31
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Hint: $$\frac{2^n}{3^{n/2}} = \left(\frac{2}{\sqrt{3}}\right)^n$$

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You should make that upper case N a lower case n. – Matt Jan 20 '11 at 4:05
Thanks! Could you list which properties did you use? – Wilbert Barrera Jan 20 '11 at 4:11
@Wilbert: You tell me. – Aryabhata Jan 20 '11 at 4:13
@MAtt: Done. Thanks. – Aryabhata Jan 20 '11 at 4:14
@Wilbert: That would be "basic arithmetic and the definition of the fractional exponents". – Arturo Magidin Jan 20 '11 at 4:19
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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 =( – Wilbert Barrera Jan 20 '11 at 4:37