Using Taylor series on an existing ODE

Question

I have had to calculate the limits for the fraction 1-cos(x) over xsin(pi x) and found the answer to be 1 over 2pi.

However, now i have to apply taylor series to the numerator and the denominator to prove the answer is correct.

 

Any help/working wold be appreciated

Frank

Answer 1

The Taylor series for 1 - cos[x] is (x² / 2) - (x⁴ / 24) + ...

 

The Taylor series for x * sin[pi x] is pi x² - pi³ x⁴ / 6 + ...

 

Since x is small, we only need to worry about the first terms in the expansions. This leaves us with just

 

lim_x->0 (x² / 2) / (pi x²) = lim_x->0 1 / (2 pi) = 1 / (2 pi)