Question

lim x-> infinity e^x / ln(1-2/x)

Answer 1

Could try L'Hopital's rule here.

Answer 2

i am suppose to do L'hospital rule..

Answer 3

Are you having trouble finding the derivatives then?

Answer 4

and then i dont even know what i was doing..i can show u my work if u want..but i dont think u can understand it ..

Answer 5

Let's just go one step at a time: L'Hopital's rule says to differentiate the top and bottom of the fraction.
Find the derivative of \(e^x\) and the derivative of \(ln(1-2/x)\).

Answer 6

yea, dont u need to first prove it is 0/0 or infinity / infinity

Answer 7

Not necessarily.

Answer 8

L'Hopital's rule will work regardless; it's just that it is most useful in those indeterminate cases.

Answer 9

Right now it is ∞/0, which is just as bad.

Answer 10

yea, i can still apply l'hopital rule even it is infinity/ 0?

Answer 11

As far as I know, L'Hopital's rule always works regardless.

Answer 12

derivative is (1/ 2-(2/x) ) (-2/x^2) / -e^2x

Answer 13

e^-x
that's becuz i bring e^x down to get 0/0

Answer 14

it's #16

Answer 15

That is not making sense to me . . .

This is what you are trying to find, right?
\(\large \lim_{x \rightarrow ∞} \frac{ e^x}{ ln(1-2/x)} \)

Answer 16

yea..

Answer 17

The derivative of \(e^x\) is \(e^x\) and the derivative of \(ln(1-2/x)\) is \(\large \frac{2x^{-2}}{1-2x^{-1}}\)

Answer 18

Does that check out to you?

Answer 19

yea, that is what i got too, it just looks messy ..

Answer 20

It can be cleaned up a little by rearranging things.
If you put the derivative of the top over the derivative of the bottom, you should be able to see that it tends to -∞

Answer 21

i actually cant see it is going to -ve infinity..

Answer 22

2/ infinity is 0 isnt it

Answer 23

It can be rearranged to \((x^2e^x-2xe^x)/2)\)

Answer 24

urgg..no idea how u do it

Answer 25

Let's see..

Derivative of e^x is e^x, derivative of ln(1-2/x) is 2/(x^2-2x). Verify these two things first.

One over the other makes (e^x)(x^2-2x)/2. You can then distribute the e^x through the parentheses.

Answer 26

isnt derivative of ln(1-2/x) = 2x^-2 / 1-2x^-1

Answer 27

Yes, but you can simplify it using rules for exponents.

Answer 28

i actually dont know how to do that..would u mind showing me how to do it :/

Answer 29

x^-2 means that it is x^2 in the denominator. so
\(\large \rightarrow \frac{2}{x^2( 1-2x^-1)}\)
You can then distribute the x^2 through the parentheses.

Answer 30

oh ok got it thx