Here's the question you clicked on:
frx
Let a>0 (a). Show using L'Hospitals rule that: \[\lim_{x \rightarrow 0} x ^{a}\ln(x)=0\] (b). By setting x=ln(t) in (a). show that: \[\lim_{x \rightarrow -\infty} |x|^{a} e ^{x}=0\] I need help with (b). I don't get the question, how do I show that (b). is true by setting x=ln(t) in (a). it doesn't make any sense to me.
the proposed substitution
\[x=\ln t\] means that \[t=e^x\]
I've got that but how can I use it?
as \[x \rightarrow -\infty, t \rightarrow ?\]
\[\lim_{x \rightarrow -\infty} \] to \[\lim_{t \rightarrow 0^+}\]
in terms of t, \[\left| x \right|^a = ?\]
that's right. and what about \[e^x\]
We could set it as just t, right?
\[\lim_{t \rightarrow 0} |\ln(t)|^{a}t\]
Because t->0 shouldn't the whole expression be equal to 0 then?
what about \[\left| \ln t \right|^a\]
\[\ln(t) \rightarrow -\infty\] but since it's the abolute value its just \[\infty \]
\[\left| \ln t \right|^a \rightarrow +\infty \]. you are correct
the transformed problem is an indeterminate form \[0*\infty \] which should be resolved either as \[\frac{ 0 }{ 0 }\] or \[\frac{ \infty }{ \infty }\]
Right, so I could tranform it to the right form and use L'Hospitals rule to get the answer correct
Should I use the form \[\frac{ t }{ |\ln(t)|^{-a} }\] and then go for L'Hospital?
how about the other form?
Just as it was, is it possible to use L'hostial on multiplication form?
lhopitals rule must be in quotient form
Which form are you thinking about then?
i'm trying your quotient form
\[-\frac{ a|\ln(t)|^{-a-1} }{ t }\] Can that be the correct derivation? If so it should give us the right answer \[\lim_{t \rightarrow 0}\frac{ 1 }{ -\frac{ a|\ln(t)|^{-a-1} }{ t } } = \frac{ 1 }{- \infty }=0\] Can this be right?
we'll look at the 2nd quotient form \[\frac{ \left| ?\ln t \right|^a }{ \frac{ 1 }{ t } }\], noting that a>0
since t->0+, \[\left| \ln t \right|=-\ln t \]
the indeterminate form is inf/inf. so we'll apply l'hop's rule once
i hate getting disconnected in the middle of the equation
\[\frac{ \frac{ -a \ln ^{a-1}(t) }{t } }{\frac{ -1 }{ t ^{2} } }\]
right. when simplified becomes \[\frac{ a(-\ln t)^{a-1} }{ 1/t }\]
here's the catch to the problem
\[(-\ln t)^a \rightarrow +\infty \] for as long as the exponent a > 0.
So it's still \[\frac{ \infty }{ \infty } \]
right. but sooner or later, the exponent becomes negative after successive applications of l'hopital's rule
\[\frac{ (a-1)a(-\ln(t))^{a-2} }{ 1/t }\]
after k applications of the rule, the indeterminate form is now \[a(a-1)(a-2)...(a-k+1)\frac{ (-\ln t)^{a-k} }{ 1/t }\] where a-k is negative
so that \[(-\ln t)^{a-k} \rightarrow 0\] because a-k is negative
the quotient form is now \[a(a-1)(a-2)...(a-k+1)\frac{ 0 }{ \infty }\] and the problem is solved!
Oh, that's clever, I get it
Thank you so much for you're help I can't thank you enought this one has been on my mind for a couple of days now!