Solving l'Hôpital's Rule: Limit x→0+ (ln(x))^x

In summary, l'Hôpital's Rule is a mathematical theorem used to evaluate limits of indeterminate forms by taking the derivative of both the numerator and denominator. It can only be used when the limit has an indeterminate form, and the process involves repeatedly taking the derivative until an answer is obtained. It is important because it allows for the evaluation of limits that would otherwise be impossible to solve, and is widely used in various fields.
  • #1
ganondorf29
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Homework Statement



Find the limit as x->0+ of (ln(x))^x

*The answer is 1*

Homework Equations



l'Hôpital's rule

The Attempt at a Solution



lim (ln(x))^x = 0^0

I took the ln of that quantity to bring down the x

lim = x*ln(ln(x))

lim = ln(ln(x)) / (1/x)

Then I used l'Hôpital's rule

1/(x*ln(x)/(1/x^2)

= 1/(x^3*ln(x))

I got stuck here. If I plug in zero I get 0^3 and a undefined answer in the denominator. Do I have to do l'Hôpital's rule on the bottom again?
 
Last edited:
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  • #2
It seems that you edited this problem while I was composing an answer. The first problem I saw was lim (-ln(x))^x, as x approached 0+.
A little while later, the minus sign in front of ln(x) was gone. Here's my work for the original problem. For the edited problem, no limit exists.

[tex]Let y = (-ln x)^x[/tex]
[tex]ln y = x ln(-ln(x)) = \frac{ln(-ln(x))}{\frac{1}{x}}[/tex]

[tex]lim_{x \rightarrow 0^+} ln y = lim_{x \rightarrow 0^+} \frac{ln(-ln(x))}{\frac{1}{x}} [\frac{\infty}{\infty}][/tex]

[tex]= lim_{x \rightarrow 0^+} \frac{\frac{1}{-ln(x)}\frac{-1}{x}}{\frac{-1}{x^2}} [/tex]
[tex]= lim_{x \rightarrow 0^+} \frac{-x}{ln(x)} = 0[/tex]
Since [tex]lim_{x \rightarrow 0^+} ln y = 0[/tex]
Then [tex]ln(lim_{x \rightarrow 0^+} y) = 0[/tex]
Or [tex]lim_{x \rightarrow 0^+} (-ln(x))^x = 1[/tex]
 

What is l'Hôpital's Rule?

L'Hôpital's Rule is a mathematical theorem that helps to evaluate limits of indeterminate forms (such as 0/0 or ∞/∞) by taking the derivative of both the numerator and denominator and then evaluating the limit again.

When can l'Hôpital's Rule be used?

L'Hôpital's Rule can only be used when the limit has an indeterminate form, meaning that both the numerator and denominator approach either 0 or infinity.

How do you solve for l'Hôpital's Rule?

To solve for l'Hôpital's Rule, take the derivative of both the numerator and denominator, and then evaluate the limit again. If the limit still has an indeterminate form, repeat the process until an answer is obtained. If the limit does not have an indeterminate form after the first application of the rule, it can be evaluated directly.

What is the limit when x approaches 0+ in the equation (ln(x))^x?

The limit when x approaches 0+ in the equation (ln(x))^x is equal to 0. This can be solved using l'Hôpital's Rule by taking the derivative of both the numerator and denominator, which results in the limit being (1/∞)^0. Since any number raised to the 0 power is equal to 1, the limit simplifies to 0.

Why is l'Hôpital's Rule important?

L'Hôpital's Rule is important because it provides a method for evaluating limits that would otherwise be impossible to solve using traditional methods. It is also a fundamental tool in calculus and is used in many applications in physics, engineering, and other fields.

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