# Solving the Limit of (lnx)^(1/(x-e)) as x Approaches e

• apoptosis
In summary, the conversation involves an individual seeking help with a math problem involving the limit of a logarithmic function with a fractional exponent. They consider using L'Hopital's rule but are unsure if it is the right approach. After receiving advice from others, they realize their mistake and are able to solve the problem. The final result is found to be e^(1/e).
apoptosis
Heya! Here's a question that I CANNOT solve for the life of me. Thanks for any direction!

## Homework Statement

lim (lnx)$$^{\frac{1}{x-e}}$$
x$$\rightarrow$$e+

^^ I'm not sure if Latext is clear, but x-e is an exponent to lnx

## Homework Equations

So, I'm thinking L'hopital's rule in starting off the question, but I'm not exactly sure if it's the right way. my work so far is in part 3...

## The Attempt at a Solution

To bring the exponent down, I bring in ln:
ln(lnx)$$\frac{1}{x-e}$$

so: $$\frac{ln(lnx)}{x-e}$$

applying l'hopitals, I find that it results in a 0/0 fraction so I derive:
to get:
$$\frac{1}{x)(lnx)(1-e)}$$
However, when I sub in e I get -2.14...so I guess I'm on the wrong track

I did graph the function, where the equation is not defined at e. As I approach e from the right, I get a number of 1.44.

Thank you once again!

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apoptosis said:
applying l'hopitals, I find that it results in a 0/0 fraction so I derive:
to get:
$$\frac{1}{x)(lnx)(1-e)}$$
However, when I sub in e I get -2.14...so I guess I'm on the wrong track

I don't think you did the derivatives right.

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However, I don't see my mistake, it'd be great if you could point it out to me.
I am following l'hopital's rule...so that's why I'm taking the derivative of the numerator and denominator separately instead of the product or quotient derivative rule.

there is no reason to apply l'hopitals rule here! l'hopitals rule is only applyed when you have a limit of the intermediate form $$\frac{0}{0}$$ or $$\frac{\infty}{\infty}$$

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

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apoptosis said:
However, I don't see my mistake, it'd be great if you could point it out to me.
I am following l'hopital's rule...so that's why I'm taking the derivative of the numerator and denominator separately instead of the product or quotient derivative rule.

Right... but what's the derivative of x-e with respect to x, unless I do not follow?

or wait is this what you meant:

$$[{ln(x)]^{\frac{1}{x-e}$$ because one cannot clearly see what u meant?
If this is the case, than you are on the right track, just double check your result after u applied l'hopitals rule!

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So, I think you are on the right track.

Let $$L=lim_{x\rightarrow e}ln(x)}^{\frac{1}{(x-e)}}$$

Then $$ln(L)=lim_{x\rightarrow e}ln(ln(x)}^{\frac{1}{(x-e)}})=lim_{x\rightarrow e}\frac{1}{x-e} ln(ln(x))$$.

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I think the problem, then, is this:

$$ln(L)=lim_{x\rightarrow e}\frac{ln(ln(x))}{(x-e)}$$

In which case, taking derivatives of the top and bottom gives:

$$\frac{d}{d x}(x-e)=1$$ (here is where you made your mistake)$$\frac{d}{d x}(ln(ln(x)))=\frac{1}{x ln(x)}$$

Then, the limit becomes $$lim_{x\rightarrow e}\frac{ln(ln(x))}{(x-e)}=lim_{x\rightarrow e}\frac{1}{x ln(x)}$$

Which is $$\frac{1}{e ln(e)}=\frac{1}{e}$$

So, $$L=e^{ln(L)}=e^{\frac{1}{e}}$$

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Aha! i see...so i just forgot to take the derivative of e's '1' exponent *smacks forehead* all this time and second guessing...

Thank you very much, everyone!

and sorry about the misunderstanding about 1/(x-e) as the exponent. still getting used to Latex. :D

## 1. What is the limit of (lnx)^(1/(x-e)) as x approaches e?

The limit of (lnx)^(1/(x-e)) as x approaches e is equal to 1. This means that as x gets closer and closer to the value of e, the expression (lnx)^(1/(x-e)) approaches 1.

## 2. How do you solve the limit of (lnx)^(1/(x-e)) as x approaches e?

To solve this limit, we can use the L'Hospital's rule. Taking the natural logarithm of both the numerator and the denominator and using the chain rule, we get ln(lnx) / (1 - 1/x). Substituting x = e, we get ln(1) / (1 - 1/e) = 0 / (1 - 1/e) = 0. Therefore, the limit is equal to 1.

## 3. What is the significance of the limit of (lnx)^(1/(x-e)) as x approaches e?

The limit of (lnx)^(1/(x-e)) as x approaches e is significant in calculus because it helps us understand the behavior of the function as it approaches a specific value. In this case, the limit approaching e is 1, which tells us that the function is approaching a horizontal asymptote at y = 1 as x approaches e.

## 4. Can we use any other method to solve this limit?

Yes, we can also use algebraic manipulation to solve this limit. By multiplying the numerator and denominator by (x-e), we get (lnx) / ((x-e)^(x-e)). As x approaches e, the expression (x-e)^(x-e) approaches 1. Therefore, the limit is equal to ln(e) = 1.

## 5. What is the graph of (lnx)^(1/(x-e)) as x approaches e?

The graph of (lnx)^(1/(x-e)) as x approaches e is a horizontal line at y = 1. This is because the limit of the function as x approaches e is 1, which means that the function approaches a horizontal asymptote at y = 1 as x gets closer and closer to e.

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