x→∞
lim x*sin2x / (x^2+1)
因为
sin2x为有界量
x/(x^2+1)=1/(x+(1/x))趋于0,为无穷小量
无穷小量乘以有界量为无穷小量
故,
lim x*sin2x / (x^2+1)=0
有不懂欢迎追问不好意思我没打清楚~~~2x/(x^2+1)是整体……x→∞ lim x * sin(2x/(x^2+1))上下同时乘以：2x/(x^2+1)=lim x * [2x/(x^2+1)] * [sin(2x/(x^2+1))] / [2x/(x^2+1)]=lim x*[2x/(x^2+1)] * lim [sin(2x/(x^2+1))] / [2x/(x^2+1)]=lim 2x^2/(x^2+1) * lim [sin(2x/(x^2+1))] / [2x/(x^2+1)]因为2x/(x^2+1)趋于0，根据重要的极限：lim sinx/x=1=lim 2x^2/(x^2+1) * 1上下同时除以x^2=lim 2x^2/x^2 / (x^2+1)/x^2=lim 2 / (1+(1/x^2))=2有不懂欢迎追问