Using L'Hospital's Rule for Solving Limits: What Are the Steps?

[KAS90]

Hey there..
I studied limits long time ago ofcourse.. but I used to use an old way in solving them, because the l'hospital rule wasn't allowed:)
My question is..can someone please help in giving me the steps I should follow in solving a limit using L'hospital's rule?
Thanks a lot..I really need to know how to solve limits..

 

[Office_Shredder]

If you're looking at the limit of f(x)/g(x), and it's in indeterminate form (either f and g go to 0, or f and g go to infinity) you can look at the limit of f'(x)/g'(x) instead, and if that limit exists it equals the limit of f/g. Sometimes you have to do this more than once

 

[HallsofIvy]

If you have a a fraction of the form f(x)/g(x) and f and g separately both go to 0 or both go to infinity (as x goes to a), then
\lim_{x\rightarrow a} \frac{f(x)}{g(x)}= \lim_{x\rightarrow a}\frac{f'(x)}{g'(x)}

That's what OfficeShredder said. I want to add that L'Hopital's rule can be used in other cases:
If we have f(x)g(x) with one of f(x) or g(x) going to 0 and the other to plus or minus infinity, then we can rewrite the problem as either f(x)/(1/g(x)) or g(x)/(1/f(x)) so we have the "0/0" or "\infty/\infty" case.

If we have F(x)= f(x)^g(x) and f(x) and g(x) both go to 0, then we can take the logarithm: ln(F(x))= g(x)ln(f(x)). Now g(x) goes to 0 while ln(f(x)) goes to negative infinity, the previous case. If this new limit is A, then the limit of F is e^A.
If we have f(x)^g(x), with f and g both going to

 

[KAS90]

Thanx a lot Office_shredder and hallsofIvy..
u guys were a lot of help..so u mean that if I don't get an indeterminate quantity, I should manipulate the functions to get an indeteminate quantity..aha..
Thanx a lot again...

 

[HallsofIvy]

No! That means if you don't get an indeterminate quantity, you don't NEED L'Hopitals rule! Just use the quantity you did get.

 

[KAS90]

oh ok..
I get it now..
Thanx again..