# L'Hospital's rule

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• February 16th 2010, 05:15 PM
Evan.Kimia
L'Hospital's rule
This may or not employ L'Hospital's, but here it is:

http://img269.imageshack.us/img269/5...00216at908.png
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.
• February 16th 2010, 05:20 PM
vince
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!(Punch))
• February 16th 2010, 05:26 PM
Drexel28
Quote:

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

http://img269.imageshack.us/img269/5...00216at908.png
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.
Quote:

Originally Posted by vince
u can apply lhopital till your ink runs dry so long as the conditions of indeterminacy are true)

Prove it! (Sun)
• February 16th 2010, 05:37 PM
Evan.Kimia
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 :(
• February 16th 2010, 05:43 PM
Evan.Kimia
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!
Quote:

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

http://img269.imageshack.us/img269/5...00216at908.png
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.

• February 16th 2010, 07:26 PM
vince
Quote:

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

Prove it! (Sun)

nice use of the sun...very cool.(Pizza)
• February 16th 2010, 07:29 PM
Prove It
Quote:

Originally Posted by Drexel28
Prove it! (Sun)

What?! (Wait)
• February 16th 2010, 07:33 PM
Drexel28
Quote:

Originally Posted by Prove It
What?! (Wait)

Haha, I actually thought of you when I said it.
• February 16th 2010, 07:37 PM
Prove It
Quote:

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 (Rofl)