设 f(0)=1 ,f'(0)= -1 求------------------ X→0时,( cosX - f(x) )÷ X =

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设 f(0)=1 ,f'(0)= -1 求------------------ X→0时,( cosX - f(x) )÷ X =

设 f(0)=1 ,f'(0)= -1 求------------------ X→0时,( cosX - f(x) )÷ X =
设 f(0)=1 ,f'(0)= -1 求------------------ X→0时,( cosX - f(x) )÷ X =

设 f(0)=1 ,f'(0)= -1 求------------------ X→0时,( cosX - f(x) )÷ X =
lim(X→0)[cosx-f(x)]/x
=lim(X→0)[-sinx-f'(x)]
=-f'(0)
=1

是不是1啊?
运用洛必达法则,因为全趋于零,所以分子分母可以求导,分子变成
-sinX-f'(X),分母变成1在X→0时(连续函数)-sin0-f'(0)= 1