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.

- anonymous

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- sirm3d

the proposed substitution

- sirm3d

\[x=\ln t\] means that \[t=e^x\]

- anonymous

I've got that but how can I use it?

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## More answers

- sirm3d

as \[x \rightarrow -\infty, t \rightarrow ?\]

- anonymous

Zero?

- anonymous

e^-infinty = 0

- sirm3d

correct

- sirm3d

so we shift from

- sirm3d

\[\lim_{x \rightarrow -\infty} \] to \[\lim_{t \rightarrow 0^+}\]

- sirm3d

in terms of t, \[\left| x \right|^a = ?\]

- anonymous

\[|\ln (t)|^{a}\] ?

- sirm3d

that's right. and what about \[e^x\]

- anonymous

We could set it as just t, right?

- sirm3d

right

- anonymous

\[\lim_{t \rightarrow 0} |\ln(t)|^{a}t\]

- anonymous

Because t->0 shouldn't the whole expression be equal to 0 then?

- sirm3d

not necessarily

- sirm3d

what about \[\left| \ln t \right|^a\]

- anonymous

\[\ln(t) \rightarrow -\infty\] but since it's the abolute value its just \[\infty \]

- sirm3d

\[\left| \ln t \right|^a \rightarrow +\infty \]. you are correct

- sirm3d

the transformed problem is an indeterminate form \[0*\infty \] which should be resolved either as \[\frac{ 0 }{ 0 }\] or \[\frac{ \infty }{ \infty }\]

- anonymous

Right, so I could tranform it to the right form and use L'Hospitals rule to get the answer correct

- anonymous

Should I use the form \[\frac{ t }{ |\ln(t)|^{-a} }\]
and then go for L'Hospital?

- sirm3d

how about the other form?

- anonymous

Just as it was, is it possible to use L'hostial on multiplication form?

- sirm3d

lhopitals rule must be in quotient form

- anonymous

Which form are you thinking about then?

- sirm3d

i'm trying your quotient form

- anonymous

\[-\frac{ a|\ln(t)|^{-a-1} }{ t }\] 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?

- sirm3d

we'll look at the 2nd quotient form
\[\frac{ \left| ?\ln t \right|^a }{ \frac{ 1 }{ t } }\], noting that a>0

- sirm3d

since t->0+, \[\left| \ln t \right|=-\ln t \]

- anonymous

Right

- sirm3d

the indeterminate form is inf/inf. so we'll apply l'hop's rule once

- sirm3d

i hate getting disconnected in the middle of the equation

- anonymous

\[\frac{ \frac{ -a \ln ^{a-1}(t) }{t } }{\frac{ -1 }{ t ^{2} } }\]

- sirm3d

right. when simplified becomes \[\frac{ a(-\ln t)^{a-1} }{ 1/t }\]

- sirm3d

here's the catch to the problem

- sirm3d

\[(-\ln t)^a \rightarrow +\infty \] for as long as the exponent a > 0.

- anonymous

So it's still \[\frac{ \infty }{ \infty } \]

- sirm3d

right. but sooner or later, the exponent becomes negative after successive applications of l'hopital's rule

- anonymous

\[\frac{ (a-1)a(-\ln(t))^{a-2} }{ 1/t }\]

- sirm3d

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

- sirm3d

so that \[(-\ln t)^{a-k} \rightarrow 0\] because a-k is negative

- sirm3d

the quotient form is now \[a(a-1)(a-2)...(a-k+1)\frac{ 0 }{ \infty }\] and the problem is solved!

- anonymous

Oh, that's clever, I get it

- anonymous

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!

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