Question

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.

Answer 1

the proposed substitution

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

Zero?

Answer 6

e^-infinty = 0

Answer 7

correct

Answer 8

so we shift from

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

We could set it as just t, right?

Answer 14

right

Answer 15

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

Answer 16

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

Answer 17

not necessarily

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

how about the other form?

Answer 25

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

Answer 26

lhopitals rule must be in quotient form

Answer 27

Which form are you thinking about then?

Answer 28

i'm trying your quotient form

Answer 29

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

Answer 30

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

Answer 31

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

Answer 32

Right

Answer 33

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

Answer 34

i hate getting disconnected in the middle of the equation

Answer 35

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

Answer 36

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

Answer 37

here's the catch to the problem

Answer 38

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

Answer 39

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

Answer 40

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

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

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

Answer 45

Oh, that's clever, I get it

Answer 46

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!