A community for students.
Here's the question you clicked on:
 0 viewing
seancashmere
 3 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)
seancashmere
 3 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)

This Question is Closed

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

hartnn
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1you get a form infinity  infinity which is undefined

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

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

kirbykirby
 3 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

seancashmere
 3 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
 3 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...

seancashmere
 3 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

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

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

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

seancashmere
 3 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
 3 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
 3 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.

seancashmere
 3 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
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1358746300788:dw

hartnn
 3 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
 3 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

seancashmere
 3 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?

seancashmere
 3 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
 3 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

seancashmere
 3 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
 3 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 ?

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

seancashmere
 3 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?

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

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

seancashmere
 3 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.

seancashmere
 3 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
 3 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.

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

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

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

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

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

hartnn
 3 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...

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

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

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

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

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

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

seancashmere
 3 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
 3 years ago
Best ResponseYou've already chosen the best response.3i am just another helper, glad i could help :) welcome ^_^
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.