Limits to infinity with natural logs

[Cacophony]

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?

 

[Punkyc7]

use Lhospital

 

[SammyS]

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.

\log_{\,b}(x^a)=(a)\log_{\,b}(x)\,.

 

[Cacophony]

I'm not allowed to use L'Hopital, and I will change #2

 

[Cacophony]

So would the second problem look something like this:

28lnx/18lnx?

 

[SammyS]

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.

 

[Cacophony]

ok so it will be ((28lnx)/x)/((18lnx)/x) = 0/0 = DNE?

 

[SammyS]

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 \displaystyle \frac{\ln(x)}{\ln(x)}\, for any x in the domain of the natural log ?

 

[Cacophony]

1/1 ?

 

[SammyS]

For #1.

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

If so then rewrite \displaystyle \frac{(\ln(x))^5}{x}\ \text{ as }\ \frac{(\ln(x))^5}{x^5}\ x^4\ .

 

[SammyS]

1/1 ?

Yes !

What does that leave you with?

 

[Cacophony]

So basically the limit is then 28/18?

 

[SammyS]

So basically the limit is then 28/18?

Yes.

 

[Cacophony]

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?

 

[Cacophony]

Ignore my last post I got confused, this isn't a derivative

 

[SammyS]

Look at post #10 .