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How do you find the taylor series of #1 + sin(x)#?

Answer 1

The general form of the Taylor series is #f(x) = Sigma_(n=0)^oo (f^(|n|) (a))/(n!)*(x-a)^n# where #f^(|n|)# denotes the #n^(th)# derivative of #f#.

Set #f(x) = sin(x)# and use #a=0#

Remember #( dsin(x))/(dx) = cos(x)# #(d cos(x))/(dx) = -sin(x)# #sin(0) = 0# #cos(0)=1#

Then the Taylor expansion of #1 + sin(x)# becomes #1 + x -(x^3)/(3!) + (x^5)/(5!) - (x^7)/(7!) + (x^9)/(9!) ...#

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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