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Limits to infinity with natural logs

  • Thread starter Cacophony
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  • #1
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Homework Statement



I'm having trouble calculating these limits.


Homework Equations



none

The Attempt at a Solution



1. lim.......((lnx)^5)/(x)=
x->infinity

(((lnx)^5)/x)/(x/x)=

(((lnx)^5)/x)/1=

How do I calculate from here?

2.lim .......(14lnx^2)/(6lnx^3)=
x->infinity

((14lnx^2)/(x^3))/((6lnx^3)/(x^3))=

((14lnx)/(x))/(6lnx)=

0/6lnx= 0

Is this right or am I forgetting something?
 

Answers and Replies

  • #2
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use Lhospital
 
  • #3
SammyS
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Homework Statement



I'm having trouble calculating these limits.


Homework Equations



none

The Attempt at a Solution



1. lim.......((lnx)^5)/(x)=
x->infinity

(((lnx)^5)/x)/(x/x)=

(((lnx)^5)/x)/1=

How do I calculate from here?
You're going in circles... back to what you started with.

As Punky said, use L'Hôpital's rule.
2.lim .......(14lnx^2)/(6lnx^3)=
x->infinity

((14lnx^2)/(x^3))/((6lnx^3)/(x^3))=

((14lnx)/(x))/(6lnx)=

0/6lnx= 0

Is this right or am I forgetting something?
In #2, what you have done is incorrect.

Use properties of logarithms for #2. It works out quite simply.

[itex]\log_{\,b}(x^a)=(a)\log_{\,b}(x)\,.[/itex]
 
  • #4
41
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I'm not allowed to use L'Hopital, and I will change #2
 
  • #5
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So would the second problem look something like this:

28lnx/18lnx?
 
  • #6
SammyS
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So would the second problem look something like this:

28lnx/18lnx?
If you mean the limit of 28lnx/(18lnx), then yes, that's right.

I'll think about #1 some more. Maybe someone else will help.
 
  • #7
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ok so it will be ((28lnx)/x)/((18lnx)/x) = 0/0 = DNE?
 
  • #8
SammyS
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ok so it will be ((28lnx)/x)/((18lnx)/x) = 0/0 = DNE?
A limit of the form 0/0 is indeterminate. It might exist or might not exist.

Why are you bothering to multiply the numerator & denominator by 1/x ? (That accomplishes nothing!)

What is [itex]\displaystyle \frac{\ln(x)}{\ln(x)}\,[/itex] for any x in the domain of the natural log ?
 
  • #10
SammyS
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For #1.

Do you know [itex] \displaystyle \lim_{x\to\infty} \frac{\ln(x)}{x}\ ?[/itex]

If so then rewrite [itex] \displaystyle \frac{(\ln(x))^5}{x}\ \text{ as }\ \frac{(\ln(x))^5}{x^5}\ x^4\ .[/itex]
 
  • #11
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  • #12
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So basically the limit is then 28/18?
 
  • #13
SammyS
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  • #14
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I think I know how to do the first one by applying the chain rule:

limit to infinity

((lnx)^5)/(x)=

(5(lnx)^4)*(1/x)/(x)=
5(lnx)^4*0/(x)= 0

Is that valid?
 
  • #15
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Ignore my last post I got confused, this isn't a derivative
 
  • #16
SammyS
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Look at post #10 .
 

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