x→∞,
lim[ 2+4+6+.+2n]/n^2=lim[(2+2n)*n/2]/n^2=lim[(2n^2+2n)/(2n^2)=1.设函数f(x)二阶可导，y=sinf(x)，求y'，y"。大师再帮我做个。用复合函数求导，得y'=cosf(x)*f'(x)，故y''=[cosf(x)*f'(x)]'=-sinf(x)*f'(x)*f'(x)+cosf(x)*f''(x)=-sinf(x)*[f'(x)]^2+cosf(x)*f''(x).