解：$(2)$∵$\frac {2x+2}{x-1}=\frac {2(x-1)+4}{x-1}=2+\frac 4{x-1}$
又∵当$x＞1$时，随着$x$的增大，$\frac 4{x-1}$的值无限接近$0$
∴$\frac {2x+2}{x-1}$的值无限接近$2$
$(3)$∵$\frac {5x-2}{x-3}=\frac {5(x-3)+13}{x-3}=5+\frac {13}{x-3}$
又∵$0≤x≤2$
∴$-13≤\frac {13}{x-3}≤-\frac {13}3$
∴$-8≤\frac {5x-2}{x-3}≤\frac 23$
