L'Hopital's Rule

[cv2yanks13]

I need to use L'Hopital's Rule... As x approaches 1.. x^1/(x-1). Am I going to have to use the logarithm technique to solve this problem? I need someone to help walk me through this problem..

 

[Deleted member 4993]

I need to use L'Hopital's Rule... As x approaches 1.. x^1/(x-1). Am I going to have to use the logarithm technique to solve this problem? I need someone to help walk me through this problem..

Please state L'Hopital's Rule finding limits with differentiation.

Problem starts from there....

 

[BigGlenntheHeavy]

$\displaystyle Let \ y \ = \ \lim_{x\to1} \ x^{\frac{1}{x-1}}$

$\displaystyle Then \ ln|y| \ = \ ln\bigg[\lim_{x\to1} \ x^{\frac{1}{x-1}}\bigg] \ = \ \lim_{x\to1} \ ln[x^{\frac{1}{x-1}}]$

$\displaystyle ln|y| \ = \ \lim_{x\to1} \ \frac{ln|x|}{x-1} \ gives \ indeterminate \ form \ \frac{0}{0}, \ ergo, \ the \ Marqui \ to \ the \ rescue.$

$\displaystyle Hence, \ ln|y| \ = \ \lim_{x\to1} \ \frac{1}{x} \ = \ 1$

$\displaystyle ln|y| \ = \ 1 \ \implies \ y \ = \ e^{1}, \ therefore \ \lim_{x\to1} \ x^{\frac{1}{x-1}} \ = \ e$