L'Hospital's Rule: Solve lim x^(1/x) as x->inf

  • Thread starter phillyolly
  • Start date
In summary, the problem can be solved using l'Hopital's rule by rewriting the function as e^lnx * (1/x) and taking the limit as x approaches infinity. This results in the form e^(lnx/x), which can be evaluated using l'Hopital's rule until a different form is obtained. Alternatively, one can use a more "artisanal" approach by defining a_n = n^(1/n) - 1 and using Newton's binomial theorem to show that the limit of a_n is 0. However, this approach was not requested by the original poster.
  • #1
phillyolly
157
0

Homework Statement

lim x^(1/x)
x->inf

The Attempt at a Solution



So, as I understand, there are several ways of solving it.
One of them, is not by using l'Hospital Rule because it is not an undeterminate function. OR IS IT?
If we rewrite the function as:
e^[lnx * (1/x)]

We get exponents: 1/infinity=0 and lnx=infinity, multiply both, it's 0.

Hence, e^0=1
 
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  • #2
Careful, you have the right answer, but you can't use infinity like a number and say 0 × ∞ = 0. If instead you had limx→∞ x/ln(x), that would be x→∞ and 1/lnx → 0, so 0 × ∞, but that limit isn't 0, it's infinity.

[tex]\lim_{x\rightarrow \infty} x^{1/x} = \lim_{x\rightarrow \infty} e^{\ln x \cdot (1/x)} = \lim_{x\rightarrow \infty} e^\frac{\ln x}{x} = e^{\lim_{x\rightarrow \infty} \frac{\ln x}{x}}[/tex]

So now you need to find lim×→∞ (lnx)/x, which you just did. :smile:
 
  • #3
phillyolly said:

Homework Statement




lim x^(1/x)
x->inf


The Attempt at a Solution



So, as I understand, there are several ways of solving it.
One of them, is not by using l'Hospital Rule because it is not an undeterminate function. OR IS IT?
This is an indeterminate form, but L'Hopital's Rule is not applicable for this particular indeterminate form.
phillyolly said:
If we rewrite the function as:
e^[lnx * (1/x)]

We get exponents: 1/infinity=0 and lnx=infinity, multiply both, it's 0.
[[itex]\infty * 0][/itex] is another indeterminate form, so you can't say with certainty that the limit of the exponent is 0.
phillyolly said:
Hence, e^0=1

Let y = ln(x)/x. Now take the natural log of both sides, and then take the limit of both sides. Presumably your text has an example that uses this approach.
 
  • #4
Bohrok said:
Careful, you have the right answer, but you can't use infinity like a number and say 0 × ∞ = 0. If instead you had limx→∞ x/ln(x), that would be x→∞ and 1/lnx → 0, so 0 × ∞, but that limit isn't 0, it's infinity.

[tex]\lim_{x\rightarrow \infty} x^{1/x} = \lim_{x\rightarrow \infty} e^{\ln x \cdot (1/x)} = \lim_{x\rightarrow \infty} e^\frac{\ln x}{x} = e^{\lim_{x\rightarrow \infty} \frac{\ln x}{x}}[/tex]

So now you need to find lim×→∞ (lnx)/x, which you just did. :smile:
Well, your last one is essentially the same as the original when lnx/x. it looks like an indeterminate function.
 
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  • #6
Thank you!
 
  • #8
phillyolly said:
Well, your last one is essentially the same as the original when lnx/x. it looks like an indeterminate function.

Well, this shouldn't be so complicated cause this is just a beginner calculus.
 
  • #9
OK, so what I do in this problem is I take a derivative twice or three times... in order to solve this indeterminate function?
 
  • #10
As long as it's 0/0 or ∞/∞, you can keep using l'Hopital's rule until you get a different form that may lead to a limit or infinity.
 
  • #11
Or for those who like it more artisanal without l'Hôpital, define:

[tex]a_n = n^{1/n} - 1[/tex]

Note that it suffices to prove that [tex]a_n \to 0[/tex].

From definition & Newton's binomium: [tex]n = (a_n + 1)^n = \sum \left( \frac{n}{i} \right) a_n^i \geq \frac{n(n-1)}{2} a_n^2^[/tex]

From which directly follows that:

[tex]\frac{2}{n-1} \geq a_n^2 \geq 0[/tex] QED

EDIT: oh just noticed how the original poster stressed the use of l'Hôpital, my apologies, this renders my post useless, wasn't aware
 
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  • #12
Bohrok said:
As long as it's 0/0 or ∞/∞, you can keep using l'Hopital's rule until you get a different form that may lead to a limit or infinity.

That is a very important note to make. That is what I was not sure of. Thank you very much.
 
  • #13
Thanks a lot for your help!

mr. vodka said:
Or for those who like it more artisanal without l'Hôpital, define:

[tex]a_n = n^{1/n} - 1[/tex]

Note that it suffices to prove that [tex]a_n \to 0[/tex].

From definition & Newton's binomium: [tex]n = (a_n + 1)^n = \sum \left( \frac{n}{i} \right) a_n^i \geq \frac{n(n-1)}{2} a_n^2^[/tex]

From which directly follows that:

[tex]\frac{2}{n-1} \geq a_n^2 \geq 0[/tex] QED

EDIT: oh just noticed how the original poster stressed the use of l'Hôpital, my apologies, this renders my post useless, wasn't aware
 

FAQ: L'Hospital's Rule: Solve lim x^(1/x) as x->inf

What is L'Hospital's Rule?

L'Hospital's Rule is a mathematical principle that allows us to evaluate certain limits involving indeterminate forms, such as 0/0 or ∞/∞. It states that if the limit of a function f(x)/g(x) as x approaches a certain value is an indeterminate form, then the limit of the derivative of f(x)/g(x) as x approaches that same value will be the same. In other words, we can take the derivative of both the numerator and denominator of a fraction and the limit will remain unchanged.

How do I apply L'Hospital's Rule?

To apply L'Hospital's Rule, you must first identify a limit that is in an indeterminate form. This typically involves dividing by 0 or approaching ∞/∞. Then, you can take the derivative of both the numerator and denominator of the fraction and evaluate the new limit. If the new limit is still in an indeterminate form, you can repeat the process until you reach a determinate form or until you can conclude that the limit does not exist.

When should I use L'Hospital's Rule?

L'Hospital's Rule is most commonly used when evaluating limits involving exponential, logarithmic, or trigonometric functions. It is also useful when evaluating limits with radicals or fractions. However, it is important to note that L'Hospital's Rule should only be used as a last resort when other methods, such as direct substitution, do not work.

What is the limit of x^(1/x) as x approaches infinity?

The limit of x^(1/x) as x approaches infinity is equal to 1. This can be evaluated using L'Hospital's Rule by taking the derivative of both the numerator and denominator, which results in the limit of (1/x)*ln(x). As x approaches infinity, ln(x) also approaches infinity, but at a slower rate, resulting in a limit of 0. Therefore, the limit of x^(1/x) as x approaches infinity is 1.

Can L'Hospital's Rule be used to evaluate limits at other values besides infinity?

Yes, L'Hospital's Rule can be used to evaluate limits at other values besides infinity. However, it is important to note that the limit must be in an indeterminate form at the specific value you are evaluating. In addition, the limit must approach that value from both the left and right sides. If these conditions are met, L'Hospital's Rule can be used to evaluate the limit at that specific value.

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