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April 23, 2014

April 23, 2014

Posted by **assignment** on Monday, March 5, 2012 at 1:08am.

- calculus -
**drwls**, Monday, March 5, 2012 at 1:24amI assume that the numerator is (x - sinx)

In a Taylor series,

x - sinx = x^3/3! - x^5/5! + ...

Divide that by x^3

The limit as x->0 is therefore 1/3! = 1/6

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