task/29999070
iim (1 - √cos2x) /x² x→ 0
" решение " (1 -√cos2*0) /0² = (1 - 1) /0 =0/0 неопредел. типа 0/0
(1 -√cos2x)/x² = (1 -√cos2x)(1+√cos2x) / x²(1+√cos2x) =
(1 - cos2x) / x²(1+√cos2x) = 2sin²(x) / x² * 1 /(1+√cos2x) =
( sin(x) / x )²* 2 / (1+√cos2x)
lim(1 -√cos2x) / x² = lim( sin(x) / x )² *lim 2/(1+√cos2x)= 1* 2/(1+1) =1.
везде x →0
ответ : 1