Question

Determine if the limit of the sequence (e^x)/(2^n) exists or diverges? Do I have to use the squeeze theorem or should I just evaluate directly?

Answer 1

I assume you mean \(n\to\infty\)

Answer 2

Yes the limit as n goes to infinity

Answer 3

\[e^x\] is just a constant then

Answer 4

do you know \[\lim_{n\to\infty}\frac{1}{2^n}\]

Answer 5

0

Answer 6

yes

Answer 7

\[\lim_{n\to\infty}\frac{e^x}{2^n}=e^x\lim_{n\to\infty}\frac{1}{2^n}\]

Answer 8

I don't understand why e^x would be a constant though considering your taking the limit as x tends to infinity. Wouldn't that mean e^x would also tend to infinity because the function e^x approaches infinity for significantly large x?

Answer 9

\[e^x\] is a constant with respect to n

Answer 10

Oh I'm sorry I gave you the wrong sequence, yah your totally right though. I meant to type the limit as n tends to infinity of e^n/2^n. I think it diverges.

Answer 11

that one does diverge

Answer 12

e>2

Answer 13

\[\frac{e^n}{2^n}=\left(\frac{e}{2}\right)^n\]

Answer 14

Okay, that makes sense, I didn't even think to rewrite it! Thank you!

Answer 15

no problem