# Limits to infinity with natural logs

1. Mar 11, 2012

### Cacophony

1. The problem statement, all variables and given/known data

I'm having trouble calculating these limits.

2. Relevant equations

none

3. 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?

2. Mar 11, 2012

### Punkyc7

use Lhospital

3. Mar 11, 2012

### SammyS

Staff Emeritus
You're going in circles... back to what you started with.

As Punky said, use L'Hôpital's rule.
In #2, what you have done is incorrect.

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

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

4. Mar 11, 2012

### Cacophony

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

5. Mar 11, 2012

### Cacophony

So would the second problem look something like this:

28lnx/18lnx?

6. Mar 11, 2012

### SammyS

Staff Emeritus
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. Mar 11, 2012

### Cacophony

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

8. Mar 11, 2012

### SammyS

Staff Emeritus
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 ?

9. Mar 11, 2012

### Cacophony

1/1 ?

10. Mar 12, 2012

### SammyS

Staff Emeritus
For #1.

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

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

11. Mar 12, 2012

### SammyS

Staff Emeritus
Yes !

What does that leave you with?

12. Mar 12, 2012

### Cacophony

So basically the limit is then 28/18?

13. Mar 12, 2012

### SammyS

Staff Emeritus
Yes.

14. Mar 12, 2012

### 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?

15. Mar 12, 2012

### Cacophony

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

16. Mar 12, 2012

### SammyS

Staff Emeritus
Look at post #10 .