lim x-> 0+ xe^(1/x) solve using L'hospital rule
lim x-> 0+ , xe^(1/x)
how you are expecting to apply L H rule here
it applies only to \[\frac{ 0 }{ 0 } n \frac{ \infty }{ \infty }\] form
then what can i do?
my chapter is all about L'hopital rule, i think there is a way to rearrange to get the limit..?
just solve it as you solve general questions of limits , see how this function behaves when x tends to zero
but i need to solve it in L' hopital rule form..if u arrange x to 1/x , then it will become 0
\[Xe^{\frac{ 1 }{ x }}\] is this your question?
yea
change it to \[\frac{ \exp \frac{ 1 }{ x } }{ \frac{ 1 }{ x } }\]
now you can differentiate
i did what sirm3d did
and then I take the ln and got -xln e / 1/x
okay you can use expansion of e^(1/x) neglect higher order term
both sides of the fraction have the factor \[-\frac{ 1 }{ x^2 }\] after differentiation, factors that you can cancel
i need to differentiate e^1/x
that would be \[e^{\frac{ 1 }{ x }} (-\frac{ 1 }{ x^2 })\] by chain rule
so i dont ln it?
seriously i dont know when to ln it and when no to
ln it when the indeterminate is 0^0,0^infinity
ok, i will try it first. thanks
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
@sirm3d srry i hve a question
yes?
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
srry i have so many questions.. just dont get these stuff ..:/
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.
this is a classic case if you don't cancel \[\frac{ -\frac{ 1 }{ x^2 } }{ \frac{ 1 }{ x^2 } }\]
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
1/0+ = +infinity
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
but if you apply L'Hop's rule, the problem wouldn't simplify
L'Hop's rule is a convenient way, NOT the best way
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
L'hop's rule may be applied several times, if it gets you somewhere, but must stop when it gets you nowhere
have to go, be back later
ok thx. see u soon
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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