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  • $\begingroup$ Okay, I see you used different code than I did for the Series and got the right result. But could you give some idea why mine didn't work? Seems like it should have. Is what you've written here just a workaround to something MMa fails at? Or did I use it wrong? Thanks for the reply. $\endgroup$ Commented Dec 31, 2023 at 19:33
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    $\begingroup$ To get the limit of f[theta] as theta -> 0 using Series, use either Series[f[theta], {theta, 0, 0}] or Series[f[theta], theta -> 0] $\endgroup$ Commented Dec 31, 2023 at 19:47
  • $\begingroup$ Right, but I'm still trying to understand why the Series[] function isn't working properly they way I used it. $\endgroup$ Commented Dec 31, 2023 at 22:59
  • $\begingroup$ Because the series you calculated is not equivalent to the limit. If you want equivalent results you need to use equivalent expressions. $\endgroup$ Commented Dec 31, 2023 at 23:35
  • $\begingroup$ The Limits I calculated above are the 0th, 1st, & 2nd derivates at theta = 0, which are the coefficients of the first three terms of a Taylor Series. So the first three terms of the Taylor Series of my function is 0 + 0 + (cot[q]/4)*theta^2. Series[] is not correctly calculating that. I'm trying to figure out how to make Series work correctly, not just find a different way to calculated the individuals limits I already calculated. Do you understand that Series[f,{x,0,n}] is supposed to give you an nth order Taylor Series? $\endgroup$ Commented Jan 1, 2024 at 0:03