Question

Compute the limit, which illustrates the use of l'Hopital's rule.
lim x^x-x / 1-x + lnx
x-->1

Answer 1

parentheses please....

Answer 2

\(\lim_{x\rightarrow 1}\frac{x^x-x}{1-x+\ln(x)}\)?

Answer 3

yes

Answer 4

you need then
\(\lim_{x\rightarrow 1}\frac{\frac{d}{dx}(x^x-x)}{\frac{d}{dx}(1-x+\ln(x))}\)


what is \(\frac{d}{dx}(x^x-x)\)?

Answer 5

x^x (ln(x) +1) -1

Answer 6

i had got the overall answer -1 but it was wrong

Answer 7

and \(\frac{d}{dx}(1-x+\ln(x))\)?

Answer 8

1/x -1

Answer 9

so we have
\(\lim_{x\rightarrow 1}\frac{x^x\ln(x)+x^x-1}{\frac{1-x}{x}}\) so again we get 0, so now we do it again

\(\frac{d}{dx}(x^x\ln(x)+x^x-1)\)?

Answer 10

so again we get 0/0 is what I meant...

Answer 11

x^x(1/x(ln(x) +1^2)

Answer 12

I get \(x^x(\frac{1}{x}+(\ln(x)+1)^2)\)

Answer 13

ya sorry thats what i meant

Answer 14

and for the other ones its -1/x^2

Answer 15

plug in 1 to the top you get 2, so -2 is the answer

Answer 16

okay thanks for some reason i had gotten -1 but i checked again now and i see that it is -2

Answer 17

dont you hate when you do all the hard stuff and it was the easy part you messed up....story of my life

Answer 18

lol agreed