求x*sin[2x/(x^2+1)]的极限

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/11 03:18:05

x是趋向于什么?不然没法求极限的
再问：不好意思，忘了写了，x趋向于无穷大。
再答： 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

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