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asked in Calculus by Veteran (14.6k points)   | 133 views

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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 (10.6k points)  
selected by
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?


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