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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)
 one year ago
 one year ago
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)
 one year ago
 one year ago

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seancashmereBest ResponseYou've already chosen the best response.0
Sorry, the condition, after I worked the problem is 1/infinity
 one year ago

hartnnBest ResponseYou've already chosen the best response.3
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.
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.1
you get a form infinity  infinity which is undefined
 one year ago

seancashmereBest ResponseYou've already chosen the best response.0
Thank you, I set it up differently :(
 one year ago

seancashmereBest ResponseYou've already chosen the best response.0
Also, aren't there other conditions for l'hospital's? For example 1^infinity?
 one year ago

hartnnBest ResponseYou've already chosen the best response.3
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.
 one year ago

hartnnBest ResponseYou've already chosen the best response.3
for L'Hopital's , only 0/0 or infinity/infinity should be there.
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.1
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
 one year ago

seancashmereBest ResponseYou've already chosen the best response.0
I got infinity over infinity when I worked that problem, how did you get zero over zero?
 one year ago

hartnnBest ResponseYou've already chosen the best response.3
i meant 0/0 or infinity / infinity ...any one of the form you can get...
 one year ago

seancashmereBest ResponseYou've already chosen the best response.0
I know, but after evaluating the above problem, I got infinity/infinity, you said you got 0/0
 one year ago

seancashmereBest ResponseYou've already chosen the best response.0
How did you work it differently?
 one year ago

hartnnBest ResponseYou've already chosen the best response.3
oh, 1/x =0, e^0 =1 11=0 <numerator 1/x=0 <denom.
 one year ago

seancashmereBest ResponseYou've already chosen the best response.0
You are right! Thank you!!
 one year ago

seancashmereBest ResponseYou've already chosen the best response.0
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.
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.1
(e^(1/x)1) / (1/x)\[\lim_{x \rightarrow \infty} \frac{e^{1/x}1}{\frac{1}{x}}= \frac{\lim_{x \rightarrow \infty}e^{1/x}1}{\lim_{x \rightarrow \infty}\frac{1}{x}}=\frac{e^{1/\infty}1}{\frac{1}{\infty}}=\frac{e^01}{0}=0/0\]
 one year ago

hartnnBest ResponseYou've already chosen the best response.3
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.
 one year ago

seancashmereBest ResponseYou've already chosen the best response.0
What? Why? I don't understand your tip at all... does infinity equal 1/0?
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.1
It's the limit as x>0 of 1/x ... it's different than just "1/x" where x=0
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.1
dw:1358746300788:dw
 one year ago

hartnnBest ResponseYou've already chosen the best response.3
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
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.1
As you go towards x>0 on the xaxis, you approach infinity because there is an asymptote at x=0. So, the graph is continuously stretching out infinitely along the asymptote
 one year ago

seancashmereBest ResponseYou've already chosen the best response.0
It's hard for me to visualize. Why is that fraction easier to use than just x as it approaches infinity?
 one year ago

seancashmereBest ResponseYou've already chosen the best response.0
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
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.1
x>0 from the right obviously, it approaches +infinity, but infinity if it approaches 0 from the left
 one year ago

seancashmereBest ResponseYou've already chosen the best response.0
1/infinity... which is zero, what do you think? Am I differentiating it correctly?
 one year ago

hartnnBest ResponseYou've already chosen the best response.3
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 ?
 one year ago

seancashmereBest ResponseYou've already chosen the best response.0
no... give me a second for my brain to process this
 one year ago

seancashmereBest ResponseYou've already chosen the best response.0
Yeah, I don't see how putting zero in a denominator will ever give me one?
 one year ago

seancashmereBest ResponseYou've already chosen the best response.0
if y is approaching zero, that is, and is in the denominator
 one year ago

hartnnBest ResponseYou've already chosen the best response.3
'putting zero in a denominator ' ??
 one year ago

seancashmereBest ResponseYou've already chosen the best response.0
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.
 one year ago

seancashmereBest ResponseYou've already chosen the best response.0
I'm supposing I actually have Zero idea what you're talking about... in the numerator
 one year ago

hartnnBest ResponseYou've already chosen the best response.3
wait, did you try differentiating without any substitution ? then you'll see what problems arise. then you put 1/x = y.
 one year ago

seancashmereBest ResponseYou've already chosen the best response.0
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.
 one year ago

hartnnBest ResponseYou've already chosen the best response.3
umm....no, you don't get 1/infinity....show your steps....
 one year ago

seancashmereBest ResponseYou've already chosen the best response.0
okay. I have to differentiate (e^(1/x)/X^2)/(1/(x^2))
 one year ago

seancashmereBest ResponseYou've already chosen the best response.0
I get e^(1/x)/(x^2) * x^2
 one year ago

seancashmereBest ResponseYou've already chosen the best response.0
that gives me e^(1/x)/x^2 * 1
 one year ago

kirbykirbyBest ResponseYou've already chosen the best response.1
(e^(1/x)/X^2)/(1/(x^2))= e^(1/x)
 one year ago

hartnnBest ResponseYou've already chosen the best response.3
numerator = e^(1/x) 1 > (1/x^2) e^(1/x) denominator =1/x > (1/x^2) overall = e^(1/x) yeah...
 one year ago

seancashmereBest ResponseYou've already chosen the best response.0
so I do that before I differentiate
 one year ago

hartnnBest ResponseYou've already chosen the best response.3
you got (1/x^2) e^(1/x) and (1/x^2) after differentiating only...
 one year ago

hartnnBest ResponseYou've already chosen the best response.3
and then you can directly put 1/x = 0, e^0=1
 one year ago

seancashmereBest ResponseYou've already chosen the best response.0
so I have 1/0, or indeterminate?
 one year ago

hartnnBest ResponseYou've already chosen the best response.3
you have 1 in the denominator now...not 0
 one year ago

seancashmereBest ResponseYou've already chosen the best response.0
got it, so my answer is 1!
 one year ago

seancashmereBest ResponseYou've already chosen the best response.0
vunderbar! so I shouldn't have differentiated in the first place.
 one year ago

seancashmereBest ResponseYou've already chosen the best response.0
I just evaluated it wrong. You are amazing. Who are you? Thanks a million!
 one year ago

hartnnBest ResponseYou've already chosen the best response.3
i am just another helper, glad i could help :) welcome ^_^
 one year ago
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