搜狗做题网求极限 lim sin(x^2 * sin (1/x))/x x->0
来源：学生作业帮 编辑：搜狗做题网作业帮 分类：数学作业 时间：2024/05/26 05:50:26
求极限 lim sin(x^2 * sin (1/x))/x x->0
求极限
lim sin(x^2 * sin (1/x))/x
x->0
求极限
lim sin(x^2 * sin (1/x))/x
x->0
∵sin(1/x)有界函数
∴lim(x->0)[xsin(1/x)]=0.(1)
∴lim(x->0)[x²sin(1/x)]=0.(2)
∵lim(x->0){sin[x²sin(1/x)]/[x²sin(1/x)]}=lim(t->0)(sint/t) (令t=x²sin(1/x),由结果(2)得)
=1 (应用重要极限).(3)
故lim(x->0){sin[x²sin(1/x)]/x}=lim(x->0){【sin[x²sin(1/x)]/[x²sin(1/x)]】*【xsin(1/x)】}
=lim(x->0){sin[x²sin(1/x)]/[x²sin(1/x)]}*lim(x->0)[xsin(1/x)]
=1*lim(x->0)[xsin(1/x)] (由结果(3)得)
=1*0 (由结果(1)得)
=0
∴lim(x->0)[xsin(1/x)]=0.(1)
∴lim(x->0)[x²sin(1/x)]=0.(2)
∵lim(x->0){sin[x²sin(1/x)]/[x²sin(1/x)]}=lim(t->0)(sint/t) (令t=x²sin(1/x),由结果(2)得)
=1 (应用重要极限).(3)
故lim(x->0){sin[x²sin(1/x)]/x}=lim(x->0){【sin[x²sin(1/x)]/[x²sin(1/x)]】*【xsin(1/x)】}
=lim(x->0){sin[x²sin(1/x)]/[x²sin(1/x)]}*lim(x->0)[xsin(1/x)]
=1*lim(x->0)[xsin(1/x)] (由结果(3)得)
=1*0 (由结果(1)得)
=0
求极限lim(x-->0)x^2 sin(1/x),
lim x->0(sqrt(sin(1/x^2)) 求极限
求极限 lim sin(x^2 * sin (1/x))/x x->0
求极限,lim x趋于0 x * sin 1/x
求极限 （（sin(x^3+x^2-x)+sin x) /x x→0 已知lim sinx/x=1
lim x趋于0 [sin(a+x)sin(a+2x)-sin^2a]/x 求极限
求极限：lim x→0 ln[1+sin^2(x)]/[e^(x^2)－1]
求极限:lim(1/sin^2(x)-1/x^2),x->0
求极限lim(x趋向于0)ln(1+x^2)/sin(1+x^2)
求极限ln(1+2x)/sin(1+2x),lim x->0
lim 1-cos4x/2sin^2x+xtan^2x x趋近于0 求极限
求极限lim(x→0) sin(1-x)/(1-x^2)