设函数f(x)={x+3,x<10,x=13x+1,x>1f(x)=\begin{cases}x + 3, & x < 1 \\ 0, & x = 1 \\ 3x + 1, & x > 1\end{cases}f(x)=⎩⎨⎧x+3,0,3x+1,x<1x=1x>1，则limx→1f(x)=\lim\limits_{x \to 1}f(x)=x→1limf(x)=_______.
Copyright ©
