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

Question

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

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?

@Wilbert: LaTeX, like you used in your first line, to make things clearer instead of relying on words like "as n goes to infinity".
Oh that was the code I used in wolfram. That's why I enclose it in a code block.
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.)

Aryabhata · Accepted Answer · 2011-01-20 04:03:08Z

up vote 8 down vote accepted

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

answered Jan 20 '11 at 4:03

Aryabhata
47.5k377166

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

4

@Wilbert: That would be "basic arithmetic and the definition of the fractional exponents". – Arturo Magidin Jan 20 '11 at 4:19

Gone · Answer 2 · 2011-01-20 05:08:04Z

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 $\rm\ \ \ a,b,m,n\ \to\ 1/3,\:2,\:2,\:n/2$

2 Answers