设函数f(x)={sin2(x−1)x−1,x<12,x=1x2−1,x>1f(x)=\begin{cases}\dfrac{\sin2(x - 1)}{x - 1},&x < 1\\2,&x = 1\\x^2 - 1,&x > 1\end{cases}f(x)=⎩⎨⎧x−1sin2(x−1),2,x2−1,x<1x=1x>1，则f(x)f(x)f(x)在x=1x = 1x=1处不连续.（ ）
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