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anonymous
 4 years 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)
anonymous
 4 years 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)

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry, the condition, after I worked the problem is 1/infinity

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.3no, 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
 4 years ago
Best ResponseYou've already chosen the best response.1you get a form infinity  infinity which is undefined

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thank you, I set it up differently :(

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

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.3if 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
 4 years ago
Best ResponseYou've already chosen the best response.3for L'Hopital's , only 0/0 or infinity/infinity should be there.

kirbykirby
 4 years ago
Best ResponseYou've already chosen the best response.1You 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
 4 years ago
Best ResponseYou've already chosen the best response.0I got infinity over infinity when I worked that problem, how did you get zero over zero?

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I know, but after evaluating the above problem, I got infinity/infinity, you said you got 0/0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0How did you work it differently?

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.3oh, 1/x =0, e^0 =1 11=0 <numerator 1/x=0 <denom.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You are right! Thank you!!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks 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
 4 years ago
Best ResponseYou've already chosen the best response.1(e^(1/x)1) / (1/x)\[\lim_{x \rightarrow \infty} \frac{e^{1/x}1}{\frac{1}{x}}= \frac{\lim_{x \rightarrow \infty}e^{1/x}1}{\lim_{x \rightarrow \infty}\frac{1}{x}}=\frac{e^{1/\infty}1}{\frac{1}{\infty}}=\frac{e^01}{0}=0/0\]

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.3one 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
 4 years ago
Best ResponseYou've already chosen the best response.0What? Why? I don't understand your tip at all... does infinity equal 1/0?

kirbykirby
 4 years ago
Best ResponseYou've already chosen the best response.1It's the limit as x>0 of 1/x ... it's different than just "1/x" where x=0

kirbykirby
 4 years ago
Best ResponseYou've already chosen the best response.1dw:1358746300788:dw

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.3x> infinity x is very very large but not equal to infinity 1/x will be very very small, but not =0 > 1/x>0

kirbykirby
 4 years ago
Best ResponseYou've already chosen the best response.1As 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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So 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
 4 years ago
Best ResponseYou've already chosen the best response.1x>0 from the right obviously, it approaches +infinity, but infinity if it approaches 0 from the left

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.01/infinity... which is zero, what do you think? Am I differentiating it correctly?

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.3see, 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
 4 years ago
Best ResponseYou've already chosen the best response.0no... give me a second for my brain to process this

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0if y is approaching zero, that is, and is in the denominator

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.3'putting zero in a denominator ' ??

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you 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
 4 years ago
Best ResponseYou've already chosen the best response.0I'm supposing I actually have Zero idea what you're talking about... in the numerator

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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
 4 years ago
Best ResponseYou've already chosen the best response.3umm....no, you don't get 1/infinity....show your steps....

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okay. I have to differentiate (e^(1/x)/X^2)/(1/(x^2))

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I get e^(1/x)/(x^2) * x^2

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that gives me e^(1/x)/x^2 * 1

kirbykirby
 4 years ago
Best ResponseYou've already chosen the best response.1(e^(1/x)/X^2)/(1/(x^2))= e^(1/x)

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.3numerator = e^(1/x) 1 > (1/x^2) e^(1/x) denominator =1/x > (1/x^2) overall = e^(1/x) yeah...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so I do that before I differentiate

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.3you got (1/x^2) e^(1/x) and (1/x^2) after differentiating only...

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.3and then you can directly put 1/x = 0, e^0=1

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so I have 1/0, or indeterminate?

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.3you have 1 in the denominator now...not 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0got it, so my answer is 1!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0vunderbar! so I shouldn't have differentiated in the first place.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I just evaluated it wrong. You are amazing. Who are you? Thanks a million!

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.3i am just another helper, glad i could help :) welcome ^_^
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