Question

\[lim_{n rightarrow infty} \frac{ e^n }{ \pi^{n/2} }\] Could someone help me a little bit here? It is a multipart question, but I think I can figure the rest of it out if I get this part. l'Hopital doesnt seem to be helping as it doesn't remove any symbols. Both denominator and the numerator goes to infinity. How should I attack the problem now?

Answer 1

\[lim_{n \rightarrow \infty} \frac{ e^n }{ \pi^{n/2} }\] Thats better.

Answer 2

isn't this just 1?
do you not end up with:
\[ \frac{\infty}{\infty} \]

Answer 3

you get \[\lim_{n \rightarrow \infty}\left( \frac{ e }{ \sqrt{\pi} } \right)^n=\]since \(e>\sqrt{\pi}\)

Answer 4

no plot shows \[+ \infty \]

Answer 5

I'm sure that is not correct :) \[ \frac{ \infty}{\infty} \] isn't the same as \[\frac{c}{c}\]

Answer 6

\(=\infty\)

Answer 7

@pgpilot326 I just have to look for the largest number to see who wins?

Answer 8

\[\pi^{\frac{n}{2}}=\pi^{\frac{1}{2}\cdot n} = (\sqrt{\pi})^n\]

Answer 9

yeah, you have basically \[\lim_{n \rightarrow \infty}r^n\]if 0<r<1, then lim goes to 0. if r>1 then lim goes to \(\infty\). if r = 1 then lim is 1

Answer 10

@pgpilot326 Ahhh of course. Your obviously right. 2.7..../(3.14...)^0.5 is more than 1 :)

Answer 11

i like it whan a plan comes together!

Answer 12

"when"

Answer 13

@pgpilot326 would an answer like e^n goes faster to infinity than pi^n/2 so it diverges to infinity also be a "good enough" answer?

Answer 14

it ultimately comes down to \(\frac{e}{\sqrt{\pi}}\)
i think it's best to rewrite the argument \[\frac{e^n}{\pi^\frac{n}{2}}\]and go from there.

Answer 15

Yeah, seems more "proper". Thanks for help!!

Answer 16

you're welcome!