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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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 Tell your teacher that it is Bernoulli's rule actually.
Last edited by courteous; January 23rd 2009 at 08:31 AM. Reason: Bernoulli
Originally Posted by courteous 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
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