下面的极限为什么不能用洛必达法则 10
Lim(x→π/2+)tanx/fan3x;Lim(x→0)(x^2*sin(1/x))/sinx题目要求是证明极限存在,但是不能用洛必达法则求就是想知道为啥说不让用洛必...
Lim(x→π/2+)tanx/fan3x;
Lim(x→0)(x^2*sin(1/x))/sinx
题目要求是证明极限存在,但是不能用洛必达法则求
就是想知道为啥说不让用洛必达法则 展开
Lim(x→0)(x^2*sin(1/x))/sinx
题目要求是证明极限存在,但是不能用洛必达法则求
就是想知道为啥说不让用洛必达法则 展开
2个回答
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Lim(x→0)(x^2*sin(1/x))/sinx
=Lim(x→0)(x/sinx)x*sin(1/x))
=Lim(x→0)x*sin(1/x) (无穷小乘以有界量)
=0
Lim(x→π/2+)tanx/tan3x
=Lim(x→π/2+)sinxcos3x/sin3xcosx
=-Lim(x→π/2+)cos3x/cosx
=-Lim(x→π/2+)sin(π/2-3x)/sin(π/2-x)
=-Lim(x→π/2+)sin(-π-3x+3π/2)/sin(π/2-x)
=-Lim(x→π/2+)sin(-3x+3π/2)/sin(π/2-x)
=-3
=Lim(x→0)(x/sinx)x*sin(1/x))
=Lim(x→0)x*sin(1/x) (无穷小乘以有界量)
=0
Lim(x→π/2+)tanx/tan3x
=Lim(x→π/2+)sinxcos3x/sin3xcosx
=-Lim(x→π/2+)cos3x/cosx
=-Lim(x→π/2+)sin(π/2-3x)/sin(π/2-x)
=-Lim(x→π/2+)sin(-π-3x+3π/2)/sin(π/2-x)
=-Lim(x→π/2+)sin(-3x+3π/2)/sin(π/2-x)
=-3
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