Can someone please explain how the taylor series would work if x, the given value from the function, is equal to a, the value at which you expand the function?(adsbygoogle = window.adsbygoogle || []).push({});

For example, let's take 1/(1-x) as an example. The taylor series for this with a=0 is Ʃ(n from 0 to infinity) x^n. But if we let x=a=0, then the first term of this sequence is 0^0, or indeterminate.

Am i supposed to take the limit of x^0 as x approaches 0-which is one in this case, or always assume x^0 to be equal to 1, or follow some other formula?

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# Indeterminate forms of Taylor series

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