128 views

asked in Calculus | 128 views

cos(x) = 1 - x2/2! + x4/4! - x6/6! + ...
sin(x) = x/1! - x3/3! + x5/5! - x7/7! + ...
cosh(x) = 1 + x2/2! + x4/4! + x6/6! + ...
sinh(x) = x/1! + x3/3! + x5/5! + x7/7! + ...
sinh(x)-sin(x)=2x3/3!+......
cosh(x)-cos(x)=2x2/2! +.....
limx->0 ( sinh(x)-sin(x) /x(cosh(x)-cos(x))
limx->0  (2x3/3!)/x*(2*x2/2! )=2/3!= 1/3

answered by Veteran (10.5k points)
selected
Why did you neglect the higher order terms in the maclaurin series expansion in the last step? Is it some kind of approximation we need do?
bcoz all of those becomes 0 during calc on x->0
Do you always follow this way with trignometric functions.
any other way to solve this instead of the series expansion?