x→∞
lim x*sin2x / (x^2+1)
因為
sin2x為有界量
x/(x^2+1)=1/(x+(1/x))趨于0,為無窮小量
無窮小量乘以有界量為無窮小量
故,
lim x*sin2x / (x^2+1)=0
有不懂歡迎追問不好意思我沒打清楚~~~2x/(x^2+1)是整體……x→∞ lim x * sin(2x/(x^2+1))上下同時乘以：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)]因為2x/(x^2+1)趨于0，根據(jù)重要的極限：lim sinx/x=1=lim 2x^2/(x^2+1) * 1上下同時除以x^2=lim 2x^2/x^2 / (x^2+1)/x^2=lim 2 / (1+(1/x^2))=2有不懂歡迎追問