跪求一解,f(x)=1/√3（x^2-3x+2）指出函数连续区间并求lim(x →0)f(x)
问题描述：
跪求一解,f(x)=1/√3（x^2-3x+2）指出函数连续区间并求lim(x →0)f(x)
答
f(x) = 1/[√3(x²-3x+2)]= 1/[√3(x-2)] - 1/[√3(x-1)]f(x)在x=1或x=2不连续∴连续区间是：(-∞,1),(1,2),(2,+∞)lim 1/[√3(x²-3x+2)] as x->0= (1/√3)lim 1/(x²-3x+2)= (1/√3)*1/(0+0+2)= 1/(2...