Question

Find the indeterminate form and locate the limit using l'hopital's rule:
Find the limit as x approaches 1 at (x^2 - 2x + 1)^x-1

Answer 1

i'm going to bed now :) but feel free to help me through the problem! i will be back before my class at 11:00am :). I will medal!!!!

Answer 2

so the problem is
\[\lim_{x \rightarrow 1}[(x^2-2x+1)^{x}-1 ]\]?

Answer 3

for some reason I think the question is to evaluate:
\[\lim_{x \rightarrow 1^+} (x^2-2x+1)^{x-1}\]
I would write x^2-2x+1 as (x-1)^2
and then I would use e^(ln(y))=y somewhere

Answer 4

Why not write it as
\[\frac{(x^2-2x+1)^{x}}{x^2-2x+1}\]then you can apply L'Hopital a few times and Bob's your uncle...

Answer 5

only nee one round of l'hospital with my method I believe

Answer 6

\[\lim_{x \rightarrow 1^+}\ln[(x-1)^{2(x-1)]} \\ =\lim_{x \rightarrow 1^+} 2(x-1) \ln(x-1) \\ =2\lim_{x \rightarrow 1^+} \frac{\ln(x-1)}{\frac{1}{x-1}}\]
apply l'hospital
then go back and do e^(L)
L being the answer to the above limit

Answer 7

lots of folks learn about L'Hopital before they know the derivative of a log...my approach only requires chain rule and polynomial derivatives.

Answer 8

he found this limit earlier actually http://openstudy.com/users/freckles#/updates/56aad602e4b0c7c29515a067

I know you didn't know that though

Answer 9

both approaches work, of course. all a matter of whether you prefer to pull the rabbit out of the hat at the beginning or the end of the trick :-)

Answer 10

I'm trying to figure out how to differentiate (x^2-2x+1)^x without knowing how to differentiate log

Answer 11

thank you both! figured it out