Solving an indeterminate 0/0 limit without L'Hospital

[Hernaner28]

Hi! I'd like to know how to solve the following limit without having to use L'Hospital's rule:

$$\displaystyle\lim_{x \to{1}}{x^{\frac{1}{x-1}}}$$

I've already solved it using the derivative rules but I want to know how to solve it without them. Thanks for your help!

EDIT: I was mistaken, it's not a 0/0 indetermiante limit, it's 1^infinity. Edit the title, thanks!

 

[scurty]

Does the limit from the left equal the limit from the right? Also, l'Hospital's rule doesn't apply to this kind of limit. What kinds does[\I] it apply to?

 

[Hernaner28]

There's one point when you're solving this limit that you get an indeterminate 0/0 limit:

$$e\displaystyle\lim_{x \to{1}}{\frac{log(x)}{x-1}}$$

When you get there you may apply L'Hospital's rule but I want to know how to continue without using it.

 

[HallsofIvy]

Well, scurty, you can use L'Hopital, indirectly: If $y= x^{1/(x-1)}$ then $ln(y)= (ln(x))/(x- 1)$which is of the form 0/0. Unfortunately, Hernaner28, I don't see any simple way to do this other than L'Hopital's rule.

 

[scurty]

Well, scurty, you can use L'Hopital, indirectly: If $y= x^{1/(x-1)}$ then $ln(y)= (ln(x))/(x- 1)$which is of the form 0/0. Unfortunately, Hernaner28, I don't see any simple way to do this other than L'Hopital's rule.

I'm confused, I thought the limit doesn't exist? The limit from the left doesn't equal the limit from the right.

Edit: I'm on a mobile phone and I completely missed the x being raised to the power, I just saw the power. The x blended in with the lim. Okay, I understand what is being asked now.

 

[Hernaner28]

Well it's alright! Maybe you can help me at this one: https://www.physicsforums.com/showthread.php?t=588011

Thanks!