Question

lim x->infinity (1-3/x)^(3x)

Answer 1

Any help? i know the answer just stuck on a certain part.

Answer 2

y = lim (1-3/x)^3x
log both sides
ln y = lim ln (1-3/x)^3x

ln y = lim 3x * ln(1-3/x)

Answer 3

Here are the terms I expanded to. Then you can see the pattern...

Answer 4

use l'hopital's rule, first you must rewrite to infinity / inifity form
ln y = lim ln(1-3/x) / (1 / 3x)

Answer 5

differentiate top and bottom, it simplifies to
ln y = lim -9 / (1 - 3/x)
now the limit is obviously -9
ln y = -9
y = e^-9 = the original limit

Answer 6

This is what i've gotten so far,
lim x-> infinity (1-3/x)^(3x)
y= (1-3/x)^(3x)
ln y = (3x) * ln(1-3/x)
lim x->infinity 3x * ln (1-3/x)
I have no idea how to do it.

Answer 7

with l'hopital's rule

Answer 8

but 3x * ln(1-3/x) is in the form Infinity * 0 so you can't do it just yet. you can move the 3x into the denominator

Answer 9

so pretty much lim x-> infinity (1-3/x)/(1/3x) then do i use L'hopitals rule?