# Math Help - L'Hospital's rule

1. ## L'Hospital's rule

This may or not employ L'Hospital's, but here it is:

Dont mind the 0 there. I know the rule means f'(x)/g'(x) as long as they are both continuous and that g'(x) doesnt equal 0. If i were to do what this rule says the first time i would get 3x^2 in my denominator which wont work since then 3(0)^2 =0. Apply it a second time and i would get...

-cos(x)-1
--------------
6x

which wont work either. If i would give it a 3rd go, would it work or am i not using the right method? Oh, and whats the derivative for -cos(x)? Thanks.

2. u can apply lhopital till your ink runs dry so long as the conditions of indeterminacy are true. in your case, one more time will do it. $\frac{d}{dx}(-\cos(x))=\sin(x)$ (YOU SHOULD KNOW THAT!)

3. Originally Posted by Evan.Kimia
This may or not employ L'Hospital's, but here it is:

Dont mind the 0 there. I know the rule means f'(x)/g'(x) as long as they are both continuous and that g'(x) doesnt equal 0. If i were to do what this rule says the first time i would get 3x^2 in my denominator which wont work since then 3(0)^2 =0. Apply it a second time and i would get...

-cos(x)-1
--------------
6x

which wont work either. If i would give it a 3rd go, would it work or am i not using the right method? Oh, and whats the derivative for -cos(x)? Thanks.
Try using Maclaurin series if you know them.
Originally Posted by vince
u can apply lhopital till your ink runs dry so long as the conditions of indeterminacy are true)
Prove it!

4. Ok so i tried going about it a 3rd time and get

sin(x)
--------
6

which still gives me 0 which isnt correct...

and i dont know the Maclaurin series

5. I got it. I wrote the 1st go wrong.

Should be

-cos(x)
-----------
3x^2

then the 2nd time...
sin(x)
---------
6x

then the 3rd time....
-cos(x)
----------
6

which gives me -1/6 (which was right).

Thanks though!
Originally Posted by Evan.Kimia
This may or not employ L'Hospital's, but here it is:

Dont mind the 0 there. I know the rule means f'(x)/g'(x) as long as they are both continuous and that g'(x) doesnt equal 0. If i were to do what this rule says the first time i would get 3x^2 in my denominator which wont work since then 3(0)^2 =0. Apply it a second time and i would get...

-cos(x)-1
--------------
6x

which wont work either. If i would give it a 3rd go, would it work or am i not using the right method? Oh, and whats the derivative for -cos(x)? Thanks.

6. Originally Posted by Drexel28
Try using Maclaurin series if you know them.

Prove it!
nice use of the sun...very cool.

7. Originally Posted by Drexel28
Prove it!
What?!

8. Originally Posted by Prove It
What?!
Haha, I actually thought of you when I said it.

9. Originally Posted by Drexel28
Haha, I actually thought of you when I said it.
You're only human - how could you NOT think of me