Need help on a limit problem

[Xylitol]

Lim (x-lnx)
x->infinite

I know that the answer is infinite because by looking at it you can tell, but I need a way to prove it using L'Hospital's rule

thanks

edit: checked the answer key and the answer is infinite, but i still don't know how to prove that the answer is infinite

 

[Juked07]

Take the derivative..

 

[Xylitol]

Originally posted by: Juked07
Take the derivative..

Lim (1-(1/x))
x->infinite

that's 1 which can't be right

i need a denominator but i can't find one that works

 

[Fenixgoon]

Originally posted by: Juked07
Take the derivative..

 

[Xylitol]

Originally posted by: Fenixgoon

Originally posted by: Juked07
Take the derivative..

that'd give me 1 for the limit which isn't right

 

[slackwarelinux]

flip the x onto the the bottom eg: (ln(x) / (x^-1))
take the derivative of the top and the bottom
flip stuff around to make it neat
take the derivative again
the limit of that should approach infinity

er. I'm a tad fuzzy, but that may work

 

[BALIstik916]

For L'Hospital's rule you would start by substituting infinity into x giving you (infinity) - (ln of infinity) = infinity - infinity which is an indeterminate case, therefore you take the derivative of both terms and get 1 - (infinity)

 

[Xylitol]

Originally posted by: slackwarelinux
flip the x onto the the bottom eg: (ln(x) / (x^-1))
take the derivative of the top and the bottom
flip stuff around to make it neat
take the derivative again
the limit of that should approach infinity

er. I'm a tad fuzzy, but that may work

wouldn't that only work if the limit were "x*ln(x)"

 

[slackwarelinux]

Originally posted by: Xylitol

Originally posted by: slackwarelinux
flip the x onto the the bottom eg: (ln(x) / (x^-1))
take the derivative of the top and the bottom
flip stuff around to make it neat
take the derivative again
the limit of that should approach infinity

er. I'm a tad fuzzy, but that may work

wouldn't that only work if the limit were "x*ln(x)"

correct, I misread the problem.

 

[Xylitol]

Originally posted by: BALIstik916
For L'Hospital's rule you would start by substituting infinity into x giving you (infinity) - (ln of infinity) = infinity - infinity which is an indeterminate case, therefore you take the derivative of both terms and get 1 - (infinity)

but the derivative of ln(x) is 1/x

am i forgetting something vital that everyone here who responded seems to know?

 

[hypn0tik]

L'Hopital's rule can be applied when you have a limit of the form 0/0 or inf/inf.

You have to try and force your problem into one of those forms.

Let y = x - ln(x)
e^y = e^(x - ln(x)) = e^x/x

lim(x-->inf) of e^y = inf/inf. Use L'Hopital's rule:
lim(x-->inf) of e^y = e^x/1 = inf.

Since, lim(x-->inf) of e^y = inf, then lim(x-->inf) of y = inf

 

[Xylitol]

Originally posted by: hypn0tik
L'Hopital's rule can be applied when you have a limit of the form 0/0 or inf/inf.

You have to try and force your problem into one of those forms.

Let y = x - ln(x)
e^y = e^(x - ln(x)) = e^x/x

lim(x-->inf) of e^y = inf/inf. Use L'Hopital's rule:
lim(x-->inf) of e^y = e^x/1 = inf.

Since, lim(x-->inf) of e^y = inf, then lim(x-->inf) of y = inf

sweet
thanks

 

[DrPizza]

Originally posted by: BALIstik916
For L'Hospital's rule you would start by substituting infinity into x giving you (infinity) - (ln of infinity) = infinity - infinity which is an indeterminate case, therefore you take the derivative of both terms and get 1 - (infinity)

That is *not* how L'Hopital's rule is applied. hypn0tik is correct.