# Thread: l'Hopital's rule

1. ## 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

2. ##  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

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

Tell your teacher that it is Bernoulli's rule actually. 3. Originally Posted by courteous $\displaystyle \lim_{x\to0}\frac{\cos\!\left(x\right)-1}{2x^2}=\lim_{x\to 0}\frac{-\sin(x)}{4x}=\text{ use L'Hopital again } = \lim_{x\to0}\frac{-\cos(x)}{4} = -\frac{1}{4}$

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

#### Search Tags

lhopital, rule 