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

Question

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

zzr0ck3r · Answer

parentheses please....

zzr0ck3r · Answer

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

anonymous · Answer

yes

zzr0ck3r · Answer

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

anonymous · Answer

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

anonymous · Answer

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

zzr0ck3r · Answer

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

anonymous · Answer

1/x -1

zzr0ck3r · Answer

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

zzr0ck3r · Answer

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

anonymous · Answer

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

zzr0ck3r · Answer

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

anonymous · Answer

ya sorry thats what i meant

anonymous · Answer

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

zzr0ck3r · Answer

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

anonymous · Answer

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

zzr0ck3r · Answer

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

anonymous · Answer

lol agreed