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)

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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)
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)

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)
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√3) 或= 1/(2√3)*√3/√3 = √3/6

连续区间:(-无穷,1),(1,2),(2,+无穷)。
lim(x →0)f(x)=√6/6