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

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- Jan 23rd 2009, 09:15 AMEmmeyh15@hotmail.coml'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

- Jan 23rd 2009, 09:29 AMcourteous
- Jan 23rd 2009, 10:03 AMJester
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