cv2yanks13
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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..
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L'Hopital's Rule
- Thread starter cv2yanks13
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cv2yanks13 said: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
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\(\displaystyle Let \ y \ = \ \lim_{x\to1} \ x^{\frac{1}{x-1}}\)
\(\displaystyle Then \ ln|y| \ = \ ln\bigg[\lim_{x\to1} \ x^{\frac{1}{x-1}}\bigg] \ = \ \lim_{x\to1} \ ln[x^{\frac{1}{x-1}}]\)
\(\displaystyle ln|y| \ = \ \lim_{x\to1} \ \frac{ln|x|}{x-1} \ gives \ indeterminate \ form \ \frac{0}{0}, \ ergo, \ the \ Marqui \ to \ the \ rescue.\)
\(\displaystyle Hence, \ ln|y| \ = \ \lim_{x\to1} \ \frac{1}{x} \ = \ 1\)
\(\displaystyle ln|y| \ = \ 1 \ \implies \ y \ = \ e^{1}, \ therefore \ \lim_{x\to1} \ x^{\frac{1}{x-1}} \ = \ e\)
\(\displaystyle Then \ ln|y| \ = \ ln\bigg[\lim_{x\to1} \ x^{\frac{1}{x-1}}\bigg] \ = \ \lim_{x\to1} \ ln[x^{\frac{1}{x-1}}]\)
\(\displaystyle ln|y| \ = \ \lim_{x\to1} \ \frac{ln|x|}{x-1} \ gives \ indeterminate \ form \ \frac{0}{0}, \ ergo, \ the \ Marqui \ to \ the \ rescue.\)
\(\displaystyle Hence, \ ln|y| \ = \ \lim_{x\to1} \ \frac{1}{x} \ = \ 1\)
\(\displaystyle ln|y| \ = \ 1 \ \implies \ y \ = \ e^{1}, \ therefore \ \lim_{x\to1} \ x^{\frac{1}{x-1}} \ = \ e\)