# L'Hospital's Rule

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

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.

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

So now you need to find lim×→∞ (lnx)/x, which you just did. Mark44
Mentor

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.
If we rewrite the function as:
e^[lnx * (1/x)]

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

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.

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

So now you need to find lim×→∞ (lnx)/x, which you just did. Well, your last one is essentially the same as the original when lnx/x. it looks like an indeterminate function.

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Thank you!

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.

OK, so what I do in this problem is I take a derivative twice or three times... in order to solve this indeterminate function?

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.

Or for those who like it more artisanal without l'Hôpital, define:

$$a_n = n^{1/n} - 1$$

Note that it suffices to prove that $$a_n \to 0$$.

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

From which directly follows that:

$$\frac{2}{n-1} \geq a_n^2 \geq 0$$ 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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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.

Thanks a lot for your help!

Or for those who like it more artisanal without l'Hôpital, define:

$$a_n = n^{1/n} - 1$$

Note that it suffices to prove that $$a_n \to 0$$.

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

From which directly follows that:

$$\frac{2}{n-1} \geq a_n^2 \geq 0$$ 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