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Mathematics 37 Online
OpenStudy (anonymous):

lim x-> 0+ xe^(1/x) solve using L'hospital rule

OpenStudy (anonymous):

lim x-> 0+ , xe^(1/x)

OpenStudy (ghazi):

how you are expecting to apply L H rule here

OpenStudy (ghazi):

it applies only to \[\frac{ 0 }{ 0 } n \frac{ \infty }{ \infty }\] form

OpenStudy (anonymous):

then what can i do?

OpenStudy (anonymous):

my chapter is all about L'hopital rule, i think there is a way to rearrange to get the limit..?

OpenStudy (ghazi):

just solve it as you solve general questions of limits , see how this function behaves when x tends to zero

OpenStudy (anonymous):

but i need to solve it in L' hopital rule form..if u arrange x to 1/x , then it will become 0

OpenStudy (ghazi):

\[Xe^{\frac{ 1 }{ x }}\] is this your question?

OpenStudy (anonymous):

yea

OpenStudy (sirm3d):

change it to \[\frac{ \exp \frac{ 1 }{ x } }{ \frac{ 1 }{ x } }\]

OpenStudy (ghazi):

now you can differentiate

OpenStudy (anonymous):

i did what sirm3d did

OpenStudy (anonymous):

and then I take the ln and got -xln e / 1/x

OpenStudy (ghazi):

okay you can use expansion of e^(1/x) neglect higher order term

OpenStudy (sirm3d):

both sides of the fraction have the factor \[-\frac{ 1 }{ x^2 }\] after differentiation, factors that you can cancel

OpenStudy (anonymous):

i need to differentiate e^1/x

OpenStudy (sirm3d):

that would be \[e^{\frac{ 1 }{ x }} (-\frac{ 1 }{ x^2 })\] by chain rule

OpenStudy (anonymous):

so i dont ln it?

OpenStudy (anonymous):

seriously i dont know when to ln it and when no to

OpenStudy (sirm3d):

ln it when the indeterminate is 0^0,0^infinity

OpenStudy (anonymous):

ok, i will try it first. thanks

OpenStudy (anonymous):

isnt e^0 = 1? well, at least for sure it is not 0

OpenStudy (anonymous):

but the denominator is 1/x m which is 0 i need to have 0/0 to use L'hopital rule

OpenStudy (anonymous):

@sirm3d srry i hve a question

OpenStudy (sirm3d):

yes?

OpenStudy (anonymous):

isnt e^0 = 1? well, at least for sure it is not 0 but the denominator is 1/x m which is 0 i need to have 0/0 to use L'hopital rule

OpenStudy (anonymous):

srry i have so many questions.. just dont get these stuff ..:/

OpenStudy (sirm3d):

e^0=1, that's for sure. when both sides of the fraction have a common factor, it's a lot easier to cancel these common factors.

OpenStudy (sirm3d):

this is a classic case if you don't cancel \[\frac{ -\frac{ 1 }{ x^2 } }{ \frac{ 1 }{ x^2 } }\]

OpenStudy (anonymous):

if i think of it this way 1/ 0+ , which 0+ suppose to be a small number, 1/ small number isnt it equal to infinity

OpenStudy (sirm3d):

1/0+ = +infinity

OpenStudy (anonymous):

but dont u get the equation u wrote when u take the derivative? i'm not suppose to take the derivative before proving it that it's 0/0 or infintiy/ infinity

OpenStudy (sirm3d):

but if you apply L'Hop's rule, the problem wouldn't simplify

OpenStudy (sirm3d):

L'Hop's rule is a convenient way, NOT the best way

OpenStudy (anonymous):

but that's what my chapter is about... well..i guess i dont need to use it since my exam is all multiple choices..but i dont know how to do it in another way ..only learned this

OpenStudy (sirm3d):

L'hop's rule may be applied several times, if it gets you somewhere, but must stop when it gets you nowhere

OpenStudy (sirm3d):

have to go, be back later

OpenStudy (anonymous):

ok thx. see u soon

OpenStudy (sirm3d):

here's an example where you apply the rule once, then you must stop. \[\frac{ \sin \frac{ 1 }{ x } }{\frac{ 1 }{ x } }\] \[x \rightarrow \infty \] got to go. be back later

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