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frx
Group Title
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.
 one year ago
 one year ago
frx Group Title
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.
 one year ago
 one year ago

This Question is Closed

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
the proposed substitution
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
\[x=\ln t\] means that \[t=e^x\]
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
I've got that but how can I use it?
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
as \[x \rightarrow \infty, t \rightarrow ?\]
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
e^infinty = 0
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
so we shift from
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
\[\lim_{x \rightarrow \infty} \] to \[\lim_{t \rightarrow 0^+}\]
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
in terms of t, \[\left x \right^a = ?\]
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
\[\ln (t)^{a}\] ?
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
that's right. and what about \[e^x\]
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
We could set it as just t, right?
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
\[\lim_{t \rightarrow 0} \ln(t)^{a}t\]
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
Because t>0 shouldn't the whole expression be equal to 0 then?
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
not necessarily
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
what about \[\left \ln t \right^a\]
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
\[\ln(t) \rightarrow \infty\] but since it's the abolute value its just \[\infty \]
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
\[\left \ln t \right^a \rightarrow +\infty \]. you are correct
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
the transformed problem is an indeterminate form \[0*\infty \] which should be resolved either as \[\frac{ 0 }{ 0 }\] or \[\frac{ \infty }{ \infty }\]
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
Right, so I could tranform it to the right form and use L'Hospitals rule to get the answer correct
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
Should I use the form \[\frac{ t }{ \ln(t)^{a} }\] and then go for L'Hospital?
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
how about the other form?
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
Just as it was, is it possible to use L'hostial on multiplication form?
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
lhopitals rule must be in quotient form
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
Which form are you thinking about then?
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
i'm trying your quotient form
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{ a\ln(t)^{a1} }{ t }\] Can that be the correct derivation? If so it should give us the right answer \[\lim_{t \rightarrow 0}\frac{ 1 }{ \frac{ a\ln(t)^{a1} }{ t } } = \frac{ 1 }{ \infty }=0\] Can this be right?
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
we'll look at the 2nd quotient form \[\frac{ \left ?\ln t \right^a }{ \frac{ 1 }{ t } }\], noting that a>0
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
since t>0+, \[\left \ln t \right=\ln t \]
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
the indeterminate form is inf/inf. so we'll apply l'hop's rule once
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
i hate getting disconnected in the middle of the equation
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{ \frac{ a \ln ^{a1}(t) }{t } }{\frac{ 1 }{ t ^{2} } }\]
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
right. when simplified becomes \[\frac{ a(\ln t)^{a1} }{ 1/t }\]
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
here's the catch to the problem
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
\[(\ln t)^a \rightarrow +\infty \] for as long as the exponent a > 0.
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
So it's still \[\frac{ \infty }{ \infty } \]
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
right. but sooner or later, the exponent becomes negative after successive applications of l'hopital's rule
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
\[\frac{ (a1)a(\ln(t))^{a2} }{ 1/t }\]
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
after k applications of the rule, the indeterminate form is now \[a(a1)(a2)...(ak+1)\frac{ (\ln t)^{ak} }{ 1/t }\] where ak is negative
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
so that \[(\ln t)^{ak} \rightarrow 0\] because ak is negative
 one year ago

sirm3d Group TitleBest ResponseYou've already chosen the best response.2
the quotient form is now \[a(a1)(a2)...(ak+1)\frac{ 0 }{ \infty }\] and the problem is solved!
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
Oh, that's clever, I get it
 one year ago

frx Group TitleBest ResponseYou've already chosen the best response.0
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!
 one year ago
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