# L'Hospital's Rule -X to the power of X

• Astrum
In summary: Yes, x = e ^ lnx and therefore x^x = e ^xlnxIn summary, the limit of x as x approaches 0 from the positive direction is equal to 1. This can be found by converting the expression to e^xlnx and evaluating the limit using L'Hopital's rule. Alternatively, it can be shown using the fact that x = e^lnx, resulting in x^x = e^xlnx.
Astrum

## Homework Statement

$Lim_{x\rightarrow0^{+}}X^{x}$

## The Attempt at a Solution

$ln(x^{x})=xln(x)=\frac{ln(x)}{x^{-1}}$

Take the derivative of the top and bottom -

$\frac{x^{-1}}{-x^{-2}}=\frac{-x^{2}}{x}$

So, we're left with: $lim_{x\rightarrow0^{+}}-x$

So, the answer would be 0, right? Where di I mess up? My book says the answer should be 1.

Since you used the ln to both sides, I think you forgot the last step to go back to the original equation and plug in the limit answer zero and hence e^0 = 1

Astrum said:

## Homework Statement

$Lim_{x\rightarrow0^{+}}X^{x}$

## The Attempt at a Solution

$ln(x^{x})=xln(x)=\frac{ln(x)}{x^{-1}}$

Take the derivative of the top and bottom -

$\frac{x^{-1}}{-x^{-2}}=\frac{-x^{2}}{x}$

So, we're left with: $lim_{x\rightarrow0^{+}}-x$

So, the answer would be 0, right? Where di I mess up? My book says the answer should be 1.

You forgot to do something. You took the ln, now to balance it out you have to raise something ;)

Ah, so, we take e to the power of - $lim_{x\rightarrow0^{+}}-x$

So you cancel out the ln?

Astrum said:
Ah, so, we take e to the power of - $lim_{x\rightarrow0^{+}}-x$

So you cancel out the ln?

No, you started with x^x and then converted it to e^xlnx and then you evaluated the limit of xlnx using lhospitals rule

so that limit was zero hence you have e^0 which is 1 hence x^x as x approaches 0 is 1.

jedishrfu said:
No, you started with x^x and then converted it to e^xlnx and then you evaluated the limit of xlnx using lhospitals rule

so that limit was zero hence you have e^0 which is 1 hence x^x as x approaches 0 is 1.

Understood. Thanks.

Another way of saying this is that you started with xx but then changed to y= ln(xx)= x ln(x). So what you found was that y[/b[ goes to 0. Because ln(x) is a continuous function that is the same as saying that ln(L)= 0 where "L" is the limit you are actually seeking. And, of course, ln(1)= 0 so L= 1.

HallsofIvy said:
Another way of saying this is that you started with xx but then changed to y= ln(xx)= x ln(x). So what you found was that y[/b[ goes to 0. Because ln(x) is a continuous function that is the same as saying that ln(L)= 0 where "L" is the limit you are actually seeking. And, of course, ln(1)= 0 so L= 1.

Well, I thought of it like this: $x=e^{ln(x)}$

This is ok, right?

Astrum said:
Well, I thought of it like this: $x=e^{ln(x)}$

This is ok, right?

Yes, x = e ^ lnx andtherefore x^x = e ^xlnx

## 1. What is L'Hospital's Rule?

L'Hospital's Rule is a mathematical theorem that helps to determine the limit of a fraction where the numerator and denominator both approach zero or infinity.

## 2. How is L'Hospital's Rule applied to functions with x to the power of x?

L'Hospital's Rule can be applied to functions with x to the power of x by taking the natural log of both the numerator and denominator and then using the limit definition of the derivative to solve for the limit.

## 3. When should L'Hospital's Rule be used?

L'Hospital's Rule should be used when evaluating a limit that results in an indeterminate form such as 0/0 or ∞/∞.

## 4. Are there any limitations to L'Hospital's Rule?

Yes, L'Hospital's Rule can only be applied when the limit is in an indeterminate form. It also cannot be used for limits that approach a non-numerical value, such as limits that approach infinity or negative infinity.

## 5. Can L'Hospital's Rule be used for functions with more than just x to the power of x?

Yes, L'Hospital's Rule can be applied to functions with other variables or exponents, as long as the limit is in an indeterminate form. However, it may become more complicated and require multiple applications of the rule.

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