Question

L'Hopital's

lim x->-inf (1-(3/x))^(2x)

Im stuck at lim """ 2xln((x-3)/x))

Answer 1

change 2x to (2x)^(-1) maybe to make this a fraction

Answer 2

Is this \[\lim_{x \rightarrow -\infty}{\frac 1{3x}}^{2x}\]?

Answer 3

\[\lim \frac{ln(1-\frac{3}{x})}{1/2x}\]

Answer 4

how do you explain that mathematically? divide by 2x^2?

Answer 5

\[\frac{1}{a}=a^{-1}\]

Answer 6

that gives you 0/(1/0) does lhop work for that?

Answer 7

1/inf*

Answer 8

its indeterminant :) but id have to chk to be sure

Answer 9

0/(1/0) is 0/infinity

Answer 10

L'hopital's only works for 0/0 or infinity/infinity

Answer 11

yeah so how do I know to stop there? I mean couldn't we manipulate it to get 0/0 or inf/inf?

Answer 12

Yes, what Amistre64 wrote does get you that

Answer 13

3/infinity is 0
ln(1)=0

Answer 14

1-3/inf = 1; ln(1) = 0

Answer 15

I get that part but can't I do something else to make it so its 0/0 or something

Answer 16

It is already 0/0

Answer 17

with what amistre wrote we have 0/ (1/-inf)) right?

Answer 18

i spose you can do anything you want to it; but why would some other thing have to be worked out if we got something that works now?

Answer 19

yes

Answer 20

1/2x = 1/inf = 0

Answer 21

I was under the impression that we determined that we can't use l'hops rule.

Answer 22

we can

Answer 23

Let me write it out shall I?

Answer 24

so it is 0/0 ahh, my apologies calc teachers here don't teach us anything very useful like we can use 1/inf lol

Answer 25

lim ln(1-3/-inf) -0
x>-inf --------- = ---- if anything :)
1/2(-inf) -0

Answer 26

this is a huge mess..