Question

How to use L'Hopital's rule on this problem?
lim x -> 0 (x/x-1 - 1/lnx)

Answer 1

please help

Answer 2

is that
\(\Large \lim \limits_{x \rightarrow 0} \dfrac{x}{x-1}-\dfrac{1}{\ln x}\)
?

Answer 3

yes

Answer 4

are you sure L'Hopital's *can* be applied ?
the form isn't 0/0 or infinity/infinity

Answer 5

i dont know, my teacher told me to use L'Hopital's rule on that problem

Answer 6

if it cant, should i put it cant on my paper?

Answer 7

L'Hopital's rule states that when you are getting an indeterminate form while evaluating a limit, then the limit can be alternatively evaluated by finding the derivative of the numerator at the value the limit approaches over the derivative of the denominator at the value the limit approaches. However, the derivative of the denominator at the value the limit approaches can't be 0 and the indeterminate form should 0/0 and both derivatives should exist. So by the rule we get:\[\lim_{x \rightarrow 0}\frac{ x \ln(x)-x+1 }{ (x-1)\ln(x)}=\lim_{x \rightarrow 0}\frac{ x }{ x-1 }-\lim_{x \rightarrow 0}\frac{ 1 }{ \ln(x) }=0- 0=0\] I really don't see how L'hoptial's rule can be applied here since we are not getting an indeterminate form. Evaluating it as a sum of limits works just fine instead.

@Nageer