求x*sin[2x/(x^2+1)]的极限

x是趋向于什么?不然没法求极限的不好意思，忘了写了，x趋向于无穷大。x→∞时2x/(x^2+1)→0所以lim(x→∞)sin[2x/(x^2+1)]/[2x/(x^2+1)]=1（等价于x→0时limsinx/x=1）lim(x→∞)2x^2/(x^2+1)=lim2/(1+1/x^2)=2所以lim(x→∞)x*sin[2x/(x^2+1)]=lim(x→∞){sin[2x/(x^2+1)]/[2x/(x^2+1)]}*[2x^2/(x^2+1)]=1*2=2