anonymous
  • anonymous
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)
Mathematics
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chestercat
  • chestercat
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anonymous
  • anonymous
Sorry, the condition, after I worked the problem is -1/infinity
hartnn
  • hartnn
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.
kirbykirby
  • kirbykirby
you get a form infinity - infinity which is undefined

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anonymous
  • anonymous
Thank you, I set it up differently :(
anonymous
  • anonymous
Also, aren't there other conditions for l'hospital's? For example 1^infinity?
hartnn
  • hartnn
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.
hartnn
  • hartnn
for L'Hopital's , only 0/0 or infinity/infinity should be there.
kirbykirby
  • kirbykirby
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
anonymous
  • anonymous
I got infinity over infinity when I worked that problem, how did you get zero over zero?
hartnn
  • hartnn
i meant 0/0 or infinity / infinity ...any one of the form you can get...
anonymous
  • anonymous
I know, but after evaluating the above problem, I got -infinity/infinity, you said you got 0/0
anonymous
  • anonymous
How did you work it differently?
hartnn
  • hartnn
oh, 1/x =0, e^0 =1 1-1=0 <--numerator 1/x=0 <--denom.
anonymous
  • anonymous
You are right! Thank you!!
anonymous
  • anonymous
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.
kirbykirby
  • kirbykirby
(e^(1/x)-1) / (1/x)\[\lim_{x \rightarrow \infty} \frac{e^{1/x}-1}{\frac{1}{x}}= \frac{\lim_{x \rightarrow \infty}e^{1/x}-1}{\lim_{x \rightarrow \infty}\frac{1}{x}}=\frac{e^{1/\infty}-1}{\frac{1}{\infty}}=\frac{e^0-1}{0}=0/0\]
hartnn
  • hartnn
ok, sure :)
anonymous
  • anonymous
Thanks Kirby!
hartnn
  • hartnn
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.
anonymous
  • anonymous
What? Why? I don't understand your tip at all... does infinity equal 1/0?
kirbykirby
  • kirbykirby
It's the limit as x->0 of 1/x ... it's different than just "1/x" where x=0
hartnn
  • hartnn
yes, it does.
kirbykirby
  • kirbykirby
|dw:1358746300788:dw|
hartnn
  • hartnn
x---> infinity x is very very large but not equal to infinity 1/x will be very very small, but not =0 -----> 1/x->0
anonymous
  • anonymous
interesting
kirbykirby
  • kirbykirby
As you go towards x->0 on the x-axis, you approach infinity because there is an asymptote at x=0. So, the graph is continuously stretching out infinitely along the asymptote
anonymous
  • anonymous
It's hard for me to visualize. Why is that fraction easier to use than just x as it approaches infinity?
anonymous
  • anonymous
So I have (-e^(1/x)/x^2)/-(1/x^2), which, using l'hospital's rule gives me 0/0. I differentiate that and I have
kirbykirby
  • kirbykirby
x->0 from the right obviously, it approaches +infinity, but -infinity if it approaches 0 from the left
anonymous
  • anonymous
1/infinity... which is zero, what do you think? Am I differentiating it correctly?
hartnn
  • hartnn
see, if you differentiate directly, you'll need to differentiate, 1/x in the denominator, which will give you -1/x^2 , which again will be 0 for x->infinity. BUT, if you put y=1/x you'll have y in the denominator, and differentiating y will give u 1. got the point ?
anonymous
  • anonymous
no... give me a second for my brain to process this
anonymous
  • anonymous
Yeah, I don't see how putting zero in a denominator will ever give me one?
anonymous
  • anonymous
if y is approaching zero, that is, and is in the denominator
hartnn
  • hartnn
'putting zero in a denominator ' ??
anonymous
  • anonymous
you said I should use y--->0 instead of 1/x as x---> infinity. If I plug y in my denominator, then I'll have zero.
anonymous
  • anonymous
I'm supposing I actually have Zero idea what you're talking about... in the numerator
hartnn
  • hartnn
wait, did you try differentiating without any substitution ? then you'll see what problems arise. then you put 1/x = y.
anonymous
  • anonymous
I tried differentiating and I wound up with 1/infinity, which is zero. I'm wondering if I got it or if I botched it.
hartnn
  • hartnn
umm....no, you don't get 1/infinity....show your steps....
anonymous
  • anonymous
okay. I have to differentiate (-e^(1/x)/X^2)/(1/(x^2))
anonymous
  • anonymous
I get -e^(1/x)/(x^2) * -x^2
hartnn
  • hartnn
wait.
anonymous
  • anonymous
that gives me -e^(1/x)/x^2 * -1
anonymous
  • anonymous
I'm waiting
kirbykirby
  • kirbykirby
(-e^(1/x)/X^2)/(1/(x^2))= -e^(1/x)
hartnn
  • hartnn
numerator = e^(1/x) -1 ---> (-1/x^2) e^(1/x) denominator =1/x ------> (-1/x^2) overall = e^(1/x) yeah...
anonymous
  • anonymous
so I do that before I differentiate
hartnn
  • hartnn
you got (-1/x^2) e^(1/x) and (-1/x^2) after differentiating only...
hartnn
  • hartnn
and then you can directly put 1/x = 0, e^0=1
anonymous
  • anonymous
so I have 1/0, or indeterminate?
hartnn
  • hartnn
you have 1 in the denominator now...not 0
anonymous
  • anonymous
got it, so my answer is 1!
hartnn
  • hartnn
yes!
anonymous
  • anonymous
e^0/1?
anonymous
  • anonymous
vunderbar! so I shouldn't have differentiated in the first place.
anonymous
  • anonymous
I just evaluated it wrong. You are amazing. Who are you? Thanks a million!
hartnn
  • hartnn
i am just another helper, glad i could help :) welcome ^_^

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