anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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sirm3d
  • sirm3d
the proposed substitution
sirm3d
  • sirm3d
\[x=\ln t\] means that \[t=e^x\]
anonymous
  • anonymous
I've got that but how can I use it?

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sirm3d
  • sirm3d
as \[x \rightarrow -\infty, t \rightarrow ?\]
anonymous
  • anonymous
Zero?
anonymous
  • anonymous
e^-infinty = 0
sirm3d
  • sirm3d
correct
sirm3d
  • sirm3d
so we shift from
sirm3d
  • sirm3d
\[\lim_{x \rightarrow -\infty} \] to \[\lim_{t \rightarrow 0^+}\]
sirm3d
  • sirm3d
in terms of t, \[\left| x \right|^a = ?\]
anonymous
  • anonymous
\[|\ln (t)|^{a}\] ?
sirm3d
  • sirm3d
that's right. and what about \[e^x\]
anonymous
  • anonymous
We could set it as just t, right?
sirm3d
  • sirm3d
right
anonymous
  • anonymous
\[\lim_{t \rightarrow 0} |\ln(t)|^{a}t\]
anonymous
  • anonymous
Because t->0 shouldn't the whole expression be equal to 0 then?
sirm3d
  • sirm3d
not necessarily
sirm3d
  • sirm3d
what about \[\left| \ln t \right|^a\]
anonymous
  • anonymous
\[\ln(t) \rightarrow -\infty\] but since it's the abolute value its just \[\infty \]
sirm3d
  • sirm3d
\[\left| \ln t \right|^a \rightarrow +\infty \]. you are correct
sirm3d
  • sirm3d
the transformed problem is an indeterminate form \[0*\infty \] which should be resolved either as \[\frac{ 0 }{ 0 }\] or \[\frac{ \infty }{ \infty }\]
anonymous
  • anonymous
Right, so I could tranform it to the right form and use L'Hospitals rule to get the answer correct
anonymous
  • anonymous
Should I use the form \[\frac{ t }{ |\ln(t)|^{-a} }\] and then go for L'Hospital?
sirm3d
  • sirm3d
how about the other form?
anonymous
  • anonymous
Just as it was, is it possible to use L'hostial on multiplication form?
sirm3d
  • sirm3d
lhopitals rule must be in quotient form
anonymous
  • anonymous
Which form are you thinking about then?
sirm3d
  • sirm3d
i'm trying your quotient form
anonymous
  • 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
  • sirm3d
we'll look at the 2nd quotient form \[\frac{ \left| ?\ln t \right|^a }{ \frac{ 1 }{ t } }\], noting that a>0
sirm3d
  • sirm3d
since t->0+, \[\left| \ln t \right|=-\ln t \]
anonymous
  • anonymous
Right
sirm3d
  • sirm3d
the indeterminate form is inf/inf. so we'll apply l'hop's rule once
sirm3d
  • sirm3d
i hate getting disconnected in the middle of the equation
anonymous
  • anonymous
\[\frac{ \frac{ -a \ln ^{a-1}(t) }{t } }{\frac{ -1 }{ t ^{2} } }\]
sirm3d
  • sirm3d
right. when simplified becomes \[\frac{ a(-\ln t)^{a-1} }{ 1/t }\]
sirm3d
  • sirm3d
here's the catch to the problem
sirm3d
  • sirm3d
\[(-\ln t)^a \rightarrow +\infty \] for as long as the exponent a > 0.
anonymous
  • anonymous
So it's still \[\frac{ \infty }{ \infty } \]
sirm3d
  • sirm3d
right. but sooner or later, the exponent becomes negative after successive applications of l'hopital's rule
anonymous
  • anonymous
\[\frac{ (a-1)a(-\ln(t))^{a-2} }{ 1/t }\]
sirm3d
  • 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
  • sirm3d
so that \[(-\ln t)^{a-k} \rightarrow 0\] because a-k is negative
sirm3d
  • sirm3d
the quotient form is now \[a(a-1)(a-2)...(a-k+1)\frac{ 0 }{ \infty }\] and the problem is solved!
anonymous
  • anonymous
Oh, that's clever, I get it
anonymous
  • 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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