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frx
 3 years ago
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.
frx
 3 years ago
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.

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sirm3d
 3 years ago
Best ResponseYou've already chosen the best response.2the proposed substitution

sirm3d
 3 years ago
Best ResponseYou've already chosen the best response.2\[x=\ln t\] means that \[t=e^x\]

frx
 3 years ago
Best ResponseYou've already chosen the best response.0I've got that but how can I use it?

sirm3d
 3 years ago
Best ResponseYou've already chosen the best response.2as \[x \rightarrow \infty, t \rightarrow ?\]

sirm3d
 3 years ago
Best ResponseYou've already chosen the best response.2\[\lim_{x \rightarrow \infty} \] to \[\lim_{t \rightarrow 0^+}\]

sirm3d
 3 years ago
Best ResponseYou've already chosen the best response.2in terms of t, \[\left x \right^a = ?\]

sirm3d
 3 years ago
Best ResponseYou've already chosen the best response.2that's right. and what about \[e^x\]

frx
 3 years ago
Best ResponseYou've already chosen the best response.0We could set it as just t, right?

frx
 3 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{t \rightarrow 0} \ln(t)^{a}t\]

frx
 3 years ago
Best ResponseYou've already chosen the best response.0Because t>0 shouldn't the whole expression be equal to 0 then?

sirm3d
 3 years ago
Best ResponseYou've already chosen the best response.2what about \[\left \ln t \right^a\]

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

sirm3d
 3 years ago
Best ResponseYou've already chosen the best response.2\[\left \ln t \right^a \rightarrow +\infty \]. you are correct

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

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

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

sirm3d
 3 years ago
Best ResponseYou've already chosen the best response.2how about the other form?

frx
 3 years ago
Best ResponseYou've already chosen the best response.0Just as it was, is it possible to use L'hostial on multiplication form?

sirm3d
 3 years ago
Best ResponseYou've already chosen the best response.2lhopitals rule must be in quotient form

frx
 3 years ago
Best ResponseYou've already chosen the best response.0Which form are you thinking about then?

sirm3d
 3 years ago
Best ResponseYou've already chosen the best response.2i'm trying your quotient form

frx
 3 years ago
Best 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?

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

sirm3d
 3 years ago
Best ResponseYou've already chosen the best response.2since t>0+, \[\left \ln t \right=\ln t \]

sirm3d
 3 years ago
Best ResponseYou've already chosen the best response.2the indeterminate form is inf/inf. so we'll apply l'hop's rule once

sirm3d
 3 years ago
Best ResponseYou've already chosen the best response.2i hate getting disconnected in the middle of the equation

frx
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ \frac{ a \ln ^{a1}(t) }{t } }{\frac{ 1 }{ t ^{2} } }\]

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

sirm3d
 3 years ago
Best ResponseYou've already chosen the best response.2here's the catch to the problem

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

frx
 3 years ago
Best ResponseYou've already chosen the best response.0So it's still \[\frac{ \infty }{ \infty } \]

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

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

sirm3d
 3 years ago
Best ResponseYou've already chosen the best response.2after 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

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

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

frx
 3 years ago
Best ResponseYou've already chosen the best response.0Oh, that's clever, I get it

frx
 3 years ago
Best ResponseYou've already chosen the best response.0Thank 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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