Question

\[\lim_{x \rightarrow \infty} {\frac{ 2^{2x} }{ 5^{x-1}}}\]

Answer 1

\[2^{2x}=(2^2)^x=4^x\] might be a good start

Answer 2

you good from there?

Answer 3

I'm thinking. I don't have a good way to deal with the exponents though.

Answer 4

make them equal

Answer 5

multiply by \(\frac{1}{5}\) and they will be equal
or if you think that is illegal, multipy top and bottom by 5, pull the top 5 out in front of the limit

Answer 6

so then I have 5* lim 4^n/5^n which would be 0

Answer 7

since 5^x is growing faster

Answer 8

yeah if you want to be real slick you could say
\[5\lim_{x\to \infty}\left(\frac{4}{5}\right)^x=0\]

Answer 9

ok,cool. Thanks!

Answer 10

yw