A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 4 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.
anonymous
 4 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.

This Question is Closed

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

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

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

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

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

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

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

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

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

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

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

anonymous
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.2\[\left \ln t \right^a \rightarrow +\infty \]. you are correct

sirm3d
 4 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 \[\frac{ \infty }{ \infty }\]

anonymous
 4 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

anonymous
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.2how about the other form?

anonymous
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.2lhopitals rule must be in quotient form

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

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

anonymous
 4 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
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.2since t>0+, \[\left \ln t \right=\ln t \]

sirm3d
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.2i hate getting disconnected in the middle of the equation

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

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

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

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

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

sirm3d
 4 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

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

sirm3d
 4 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
 4 years ago
Best ResponseYou've already chosen the best response.2so that \[(\ln t)^{ak} \rightarrow 0\] because ak is negative

sirm3d
 4 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!

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

anonymous
 4 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!
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.