Question

L'Hopitals rule help?

Answer 1

lim (x-> infinity) (1+a/x)^bx

Answer 2

this is a 1^inf huh..

Answer 3

Yep

Answer 4

let y equal that so

lny = (bx) ln (1 + a/x)

so now you have (inf)(0)

[ln(1+a/x)]/(1/bx)

makes it 0/0

Answer 5

now you can do le l'hospital magic

Answer 6

but dont forget that after you l'hospital it..it's just ln y...you need y

Answer 7

My working so far

Let y = (1+a/x)^bx
lny = bx ln (1+a/x)

lim (x-> infinity) ln y = lim (x-> infinity) (bx ln(1+a/x)
= lim (x-> infinity) d/dx [bx * ln (1+a/x)
= lim (x-> infinity) [b * ln(1+a/x) + ln(x)/(1+a/x) + bx
= b*0 + infinity + inifinity
= infinity

Answer 8

But I am not sure as to whether ln(infinity) is infinity as I worked out?

Answer 9

you derived early...it's not yet in the form 0/0

Answer 10

But it was in the form of infinity * 0 wasn't it?

Answer 11

you can only use l'hospital when in the form 0/0 or inf/inf

Answer 12

no lgb bles is right

Answer 13

lim (x-> infinity) ln y = lim (x-> infinity) (bx ln(1+a/x)

So where do I go from there?

Answer 14

lgbasallote, I tried doing it like this

lim (x-> infinity) ln(1+a/x)/(1/bx)
lim (x-> infinity) [ln x /(1+a/x)]\ln(bx) x b
ln(infinity) /ln(infinity) * b

= infinity/infinity?

Answer 15

hmmm then l'hospital the second line again...

Answer 16

Ah ok, thanks. I have a lecture now, I'll be back in an hour