设函数f(x)在x=0处连续，且lim（h→0）f（h^2）/h^2=1,为什么f(0)=0

Question

如题所述

Answer 1

简单分析一下，答案如图所示

Answer 2

用柱坐标 F(t)=∫ ∫ ∫ (k)[x^2+f(x^2+y^2)]dxdydz =∫ ∫ ∫ [r2cos2θ+f(r2)]rdzdrdθ =∫[0--->2π]dθ ∫ [0--->t]dr∫[0--->h] [r2cos2θ+f(r2)]rdz =h∫[0--->2π]dθ ∫[0--->t] [r2cos2θ+f(r2)]rdr =h∫[0--->2π]dθ ∫[0--->t] r3cos2θdr+h∫[0--->2π]dθ ∫[0--->t] rf(r2)dr =(h/2)∫[0--->2π] (1+cos2θ)dθ ∫[0--->t] r3dr+h∫[0--->2π]dθ ∫[0--->t] rf(r2)dr =(h/2)(θ+(1/2)sin2θ) |[0--->2π] ∫[0--->t] r3dr+h∫[0--->2π]dθ ∫[0--->t] rf(r2)dr =πh∫[0--->t] r3dr+h∫[0--->2π]dθ ∫[0--->t] rf(r2)dr 因此：F'(t)=πht3+2πh*t f(t2)本回答被网友采纳