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How do you find the Maclaurin Series for #f(x)= (sinx)/x#?

Answer 1

Christopher Agresta

#(sinx)/x = sum_(n=0)^oo (-1)^n x^(2n)/((2n+1)!)#

Consider the MacLaurin series for #sinx#:

#sinx = sum_(n=0)^oo (-1)^n x^(2n+1)/((2n+1)!)#

and divide by #x# term by term:

#(sinx)/x = sum_(n=0)^oo (-1)^n 1/x x^(2n+1)/((2n+1)!) = sum_(n=0)^oo (-1)^n x^(2n)/((2n+1)!)#

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Answer 2

Axel Alvarez

To find the Maclaurin series for ( f(x) = \frac{\sin(x)}{x} ), you can start by using the known Maclaurin series expansion for ( \sin(x) ), which is ( \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots ). Then, divide each term of this series by ( x ) to get the Maclaurin series for ( \frac{\sin(x)}{x} ). The resulting series will be ( 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \cdots ).

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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.