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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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\[\lim_{x \rightarrow 0} cosx-1/x\]
cos(x-1/x)?
(cosx-1)/x

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Other answers:

I had my answer ready for the other one, I need to think about this one.
are you allowed to use l'Hospitals rule?
nope. Professor said that if we use it he won't consider the answer
well this limit is very well-known, and is usually proven geometrically, so... I'm not sure what we want to do here
there are 2 ways to finding the answer, and I know it's 0. I know one of them, wich takes too long, i want the other.
You can write cos(x) as a series, that'll work.
^that is true, that would be the only other way to prove it besides the geometric way which is given here: http://ocw.mit.edu/courses/mathematics/18-01sc-single-variable-calculus-fall-2010/part-a-definition-and-basic-rules/session-7-derivatives-of-sine-and-cosine/
so there are technically 3 ways of finding the answer
yep. MIT always saving my life. Thanks again :P
that is the long way though^ the "short" way is l'Hospitals rule
but you're welcome :)

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