Question

More help with l'hospital's rule: can I evaluate a problem using l'hospital's rule when the following condition exists: lim(x-->infinity) (xe^(1/x)-x)

Answer 1

Sorry, the condition, after I worked the problem is -1/infinity

Answer 2

no, L'Hopital's can be applied only if you have the form 0/0 or \(\infty\)/\(\infty\)
so, what you can do is,
(xe^(1/x)-x)
=x( (e^(1/x)-1))
= (e^(1/x)-1) / (1/x)
now its 0/0 form and you can apply L'Hopitals.

Answer 3

you get a form infinity - infinity which is undefined

Answer 4

Thank you, I set it up differently :(

Answer 5

Also, aren't there other conditions for l'hospital's? For example 1^infinity?

Answer 6

if you have 1^infinity form, you can apply log to both sides to get 0/0 form, but there are other ways to deal with 1^infinity form also.

Answer 7

for L'Hopital's , only 0/0 or infinity/infinity should be there.

Answer 8

You can only apply L'Hopital's rule when you have the form \[0/0\] or \[\infty/\infty\]. When you have 1^infinity, you usually transform your by using ln (natural log) on both sides of your equation (set your limit as y = lim... blabla) then ln both sides

Answer 9

I got infinity over infinity when I worked that problem, how did you get zero over zero?

Answer 10

i meant 0/0 or infinity / infinity ...any one of the form you can get...

Answer 11

I know, but after evaluating the above problem, I got -infinity/infinity, you said you got 0/0

Answer 12

How did you work it differently?

Answer 13

oh, 1/x =0,
e^0 =1
1-1=0 <--numerator
1/x=0 <--denom.

Answer 14

You are right! Thank you!!

Answer 15

Thanks for the help. Keep your eyes peeled, I may need help differentiating it. Soon. I'll close this for now and open a new one if I need help differentiating it.

Answer 16

(e^(1/x)-1) / (1/x)\[\lim_{x \rightarrow \infty} \frac{e^{1/x}-1}{\frac{1}{x}}= \frac{\lim_{x \rightarrow \infty}e^{1/x}-1}{\lim_{x \rightarrow \infty}\frac{1}{x}}=\frac{e^{1/\infty}-1}{\frac{1}{\infty}}=\frac{e^0-1}{0}=0/0\]

Answer 17

ok, sure :)

Answer 18

Thanks Kirby!

Answer 19

one tip,
before differentiating, substitute x =1/y
so, that y-->0
then diff. num and denom. seperately.
it'll much easier than to diff. by x.

Answer 20

What? Why? I don't understand your tip at all... does infinity equal 1/0?

Answer 21

It's the limit as x->0 of 1/x ... it's different than just "1/x" where x=0

Answer 22

yes, it does.