$$ \lim\limits_{n\to\infty}\frac{2^n}{3^{\frac{n}{2}}}$$
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?
$\endgroup$ 62 Answers
$\begingroup$Hint: $$\frac{2^n}{3^{n/2}} = \left(\frac{2}{\sqrt{3}}\right)^n$$
$\endgroup$ 6 $\begingroup$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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