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

 

[jhicks]

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.

 

[apoptosis]

Hi, thanks for the reply.
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.

 

[sutupidmath]

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 $$\frac{\infty}{\infty}$$

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

 

[jhicks]

Hi, thanks for the reply.
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?

 

[sutupidmath]

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!

 

[Doom of Doom]

So, I think you are on the right track.

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

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

 

[Doom of Doom]

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

 

[apoptosis]

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