# Limits to infinity with natural logs

## Homework Statement

I'm having trouble calculating these limits.

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?

Related Calculus and Beyond Homework Help News on Phys.org
use Lhospital

SammyS
Staff Emeritus
Homework Helper
Gold Member

## Homework Statement

I'm having trouble calculating these limits.

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.

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

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

So would the second problem look something like this:

28lnx/18lnx?

SammyS
Staff Emeritus
Homework Helper
Gold Member
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.

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

SammyS
Staff Emeritus
Homework Helper
Gold Member
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 ?

1/1 ?

SammyS
Staff Emeritus
Homework Helper
Gold Member
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
Staff Emeritus
Homework Helper
Gold Member
1/1 ?
Yes !

What does that leave you with?

So basically the limit is then 28/18?

SammyS
Staff Emeritus
Homework Helper
Gold Member
So basically the limit is then 28/18?
Yes.

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?

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

SammyS
Staff Emeritus
Homework Helper
Gold Member
Look at post #10 .