Question

lim of 6(x!)/(-4)^x?

Answer 1

lim of 6(x!)/(-4)^x as x approaches what?

Answer 2

infinity

Answer 3

\[\lim_{a \rightarrow \infty}6x!/-4^x=\infty\]

Answer 4

x*

Answer 5

it is not right

Answer 6

According to wolfram alpha, it's complex infinity :/

http://www.wolframalpha.com/input/?i=lim+x+approaches+infinity+6%28x%21%29%2F%28-4%29%5Ex

Because the limit as x=> inf. of (-4)^x is complex infinity. I've no idea how to prove that limit though...

Answer 7

that series expansion does look complex

Answer 8

So the limit does not exist?

Answer 9

Okay it does not exist but I don't know why

Answer 10

It's complex infinity because you don't know whether the limit is +infinity or -infinity because of the denominator, all you know is that the modulus of the limit on the Wessel (Argand) plane is infinite.

Answer 11

That is, is \[\lim_{x \rightarrow \infty}(-4)^x= +\infty \text{ or} - \infty?\]