# l'Hopital's rule

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• January 23rd 2009, 08:15 AM
Emmeyh15@hotmail.com
l'Hopital's rule
Find the limit by using the l' Hopital's rule!
\lim_{x\to0}\frac{\cos\!\left(x\right)-1}{2x^2}

So far i got to
\lim_{x\to0}\frac{\-sin\!\left(x}{4x}

What do you do next

• January 23rd 2009, 08:29 AM
courteous
Quote:

Originally Posted by Emmeyh15@hotmail.com
Find the limit by using the l' Hopital's rule!
\lim_{x\to0}\frac{\cos\!\left(x\right)-1}{2x^2}

So far i got to
\lim_{x\to 0}\frac{\-sin\!\left(x}{4x}

What do you do next

$\lim_{x\to0}\frac{\cos\!\left(x\right)-1}{2x^2}=\lim_{x\to 0}\frac{-\sin(x)}{4x}=\text{ use L'Hopital again } = \lim_{x\to0}\frac{-\cos(x)}{4} = -\frac{1}{4}$

Tell your teacher that it is Bernoulli's rule actually. (Smirk)
• January 23rd 2009, 09:03 AM
Jester
Quote:

Originally Posted by courteous
$\lim_{x\to0}\frac{\cos\!\left(x\right)-1}{2x^2}=\lim_{x\to 0}\frac{-\sin(x)}{4x}=\text{ use L'Hopital again } = \lim_{x\to0}\frac{-\cos(x)}{4} = -\frac{1}{4}$

Tell your teacher that it is Bernoulli's rule actually. (Smirk)

But is you call it Bernoulli's rule most (if not all) people won't know what your talking about.

Here's an interesting site.

Calculus I Notes, Section 4-5