Question

间断点的判断设函数f(x)= 2+（x-1)sin1/x-1 (x＜1) 2x^2+lnx (x ≥1)则x=1是f(x

Answer 1

连续点 lim(x->1-)f(x) =lim(x->1-)[2+（x-1)sin1/x-1 ] =2+0 =2, lim(x->1+)f(x) =lim(x->1+)(2x^2+lnx) =2+ln1 =2+0 =2, f(1)=2*1^2+ln1=2, 因为lim(x->1-)f(x)=f(1)=lim(x->1+)f(x), 所以f(x)在x=1点连续，即 x=1是f(x)的连续点。