Limits to infinity with natural logs

You need to use the fact that \ln(x^a)=a\ln(x).Then you have\displaystyle \lim_{x\to\infty} \frac{(\ln(x))^5}{x} = \lim_{x\to\infty} \frac{5(\ln(x))^4}{x^2} = \lim_{x\to\infty} \frac{20(\ln(x))^3}{x^3} = \lim_{x\to\infty} \frac{60(\ln(x))^2}{x^4} = \lim_{x\to\infty} \frac{120\ln(x)}{x^5} = \lim
  • #1
Cacophony
41
0

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?
 
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  • #2
use Lhospital
 
  • #3
Cacophony said:

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
I'm not allowed to use L'Hopital, and I will change #2
 
  • #5
So would the second problem look something like this:

28lnx/18lnx?
 
  • #6
Cacophony said:
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
ok so it will be ((28lnx)/x)/((18lnx)/x) = 0/0 = DNE?
 
  • #8
Cacophony said:
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 ?
 
  • #9
1/1 ?
 
  • #10
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
Cacophony said:
1/1 ?
Yes !

What does that leave you with?
 
  • #12
So basically the limit is then 28/18?
 
  • #13
Cacophony said:
So basically the limit is then 28/18?
Yes.
 
  • #14
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
Ignore my last post I got confused, this isn't a derivative
 
  • #16
Look at post #10 .
 

Related to Limits to infinity with natural logs

1. What is the limit of ln(x) as x approaches infinity?

The limit of ln(x) as x approaches infinity is equal to infinity. This means that as x gets larger and larger, the natural logarithm of x also gets larger and larger without bound.

2. How do you find the limit of a natural logarithm as the input approaches infinity?

To find the limit of a natural logarithm as the input approaches infinity, you can use the rule that states ln(x) approaches infinity as x approaches infinity. You can also use L'Hôpital's rule to evaluate the limit if the natural logarithm is part of a more complex function.

3. Can the natural logarithm approach a finite limit as the input goes to infinity?

No, the natural logarithm cannot approach a finite limit as the input goes to infinity. This is because the logarithm function is unbounded and continues to increase without bound as the input gets larger.

4. What is the limit of ln(x) as x approaches negative infinity?

The limit of ln(x) as x approaches negative infinity is undefined. This is because the natural logarithm is only defined for positive inputs, so it does not make sense to evaluate the limit as x approaches negative infinity.

5. Can the natural logarithm approach a negative infinity limit as the input goes to infinity?

Yes, the natural logarithm can approach a negative infinity limit as the input goes to infinity. This is because as x approaches infinity, ln(x) also approaches infinity, but in the negative direction. Therefore, the limit is equal to negative infinity.

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