limits problem

Question

lim   1/x⁸  [1-cos(x²/2)] [1-cos(x²/4)] =

x→0

 

(i) 1/8

(ii)1/8²

(iii)2/8³

(iv)2/8⁴

 

Answer 1

One way to do this is to apply L'Hopital's rule several times until the limit is no longer 0/0. Another easier way is to take the first two terms of MacLaurin's series for the cosine terms in the numerator and multiply these together, then divide out the denominator.

 

 

The MacLaurin series  for cosine is cos(z) = (1 - z^2/2! +z^4/4! -.....). Now multiply ((1 - z^2/2!)*((1 - z^2/2!)

 

Now let substitute x^2/2 for every z.  Divide out x^8 and let x go to zero.