Question

求问高数极限题f(x)在x=1可导 f(1)=0 f(1)'=2 求 f((sinx)^2 + cosx)/xtanx

Answer 1

lim(x→0) f((sinx)^2＋cosx)/xtanx

＝lim(x→0) f((sinx)^2＋cosx) / x^2

＝lim(x→0) [f(1－(cosx)^2＋cosx)－f(1)] / x^2

＝lim(x→0) [f(1－(cosx)^2＋cosx)－f(1)] / [－(cosx)^2＋cosx] × [－(cosx)^2＋cosx]/x^2

＝f'(1)×lim(x→0) [－(cosx)^2＋cosx]/x^2

＝2×lim(x→0) [－cosx ＋ 1]/x^2

＝2×lim(x→0) [1/2×x^2]/x^2

＝2×1/2＝1