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

you get a form infinity - infinity which is undefined

Thank you, I set it up differently :(

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

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

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

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

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

How did you work it differently?

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

You are right! Thank you!!

ok, sure :)

Thanks Kirby!

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

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

yes, it does.

|dw:1358746300788:dw|

interesting

1/infinity... which is zero, what do you think? Am I differentiating it correctly?

no... give me a second for my brain to process this

Yeah, I don't see how putting zero in a denominator will ever give me one?

if y is approaching zero, that is, and is in the denominator

'putting zero in a denominator ' ??

I'm supposing I actually have Zero idea what you're talking about... in the numerator

umm....no, you don't get 1/infinity....show your steps....

okay. I have to differentiate (-e^(1/x)/X^2)/(1/(x^2))

I get -e^(1/x)/(x^2) * -x^2

wait.

that gives me -e^(1/x)/x^2 * -1

I'm waiting

(-e^(1/x)/X^2)/(1/(x^2))= -e^(1/x)

so I do that before I differentiate

you got (-1/x^2) e^(1/x) and (-1/x^2) after differentiating only...

and then you can directly put 1/x = 0, e^0=1

so I have 1/0, or indeterminate?

you have 1 in the denominator now...not 0

got it, so my answer is 1!

yes!

e^0/1?

vunderbar! so I shouldn't have differentiated in the first place.

I just evaluated it wrong. You are amazing. Who are you? Thanks a million!

i am just another helper, glad i could help :)
welcome ^_^