1-cos2x除以xsinx
@边叛5195：lim(x→0)(1 - cos2x)÷(xsinx) -
范瑶19563289551…… lim(x趋向0)(1-cos2x)/xsinx =lim(x趋向0)[(1-1+2Sin^2(x)] /xsinx =lim(x趋向0)2sin^2x/xsinx =lim(x趋向0)2sinx/x =2
@边叛5195：lim(x→无穷)(1 - cos2x)/xsinx 有木有解,如何解? -
范瑶19563289551…… 有解.1-cos2x=2(sinx)^2, lim(x→无穷)(1-cos2x)/xsinx=lim(x→无穷)(2(sinx)^2)/xsinx=2lim(x→无穷)sinx/x, 因为|sinx|<=1,是有界变量,lim(x→无穷)1/x=0.是无穷小量,根据有界变量乘无穷小量仍为有界变量,得原式的极限为0
@边叛5195：x趋近于0 求(1 - cos2x)/(xsinx) 的极限 -
范瑶19563289551…… 解: lim【x→0】(1-cos2x)/(xsinx) =lim【x→0】2sin²x/(xsinx) =lim【x→0】(2sinx)/x =2 答案:2
@边叛5195：x趋近于0 求(1 - cos2x)/(xsinx) 的极限 - 作业帮
范瑶19563289551…… [答案] lim【x→0】(1-cos2x)/(xsinx) =lim【x→0】2sin²x/(xsinx) =lim【x→0】(2sinx)/x =2 答案:2
@边叛5195：求lim(x→∞)1 - cos2x/xsinx -
范瑶19563289551…… 1-cos2x=1-(1-2*(sinX)^2)=2*(sinX)^2,1-cos2x/xsinx=2sinx/x 因为当x→∞时,1/x→0 又sinx为有界函数,|sinx|≤1 所以lim【x→∞】sinx/x=0 lim(x→∞)1-cos2x/xsinx=0
@边叛5195：求极限limx 0时1 - cos2x/xsinx等于多少 -
范瑶19563289551…… 1-cos2x=2(sinx)^2利用等价无穷小式子=2x^2/x^2=2
@边叛5195：求lim(x→∞)1 - cos2x/xsinx - 作业帮
范瑶19563289551…… [答案] 1-cos2x=1-(1-2*(sinX)^2)=2*(sinX)^2,1-cos2x/xsinx=2sinx/x 因为当x→∞时,1/x→0 又sinx为有界函数,|sinx|≤1 所以lim【x→∞】sinx/x=0 lim(x→∞)1-cos2x/xsinx=0
@边叛5195：limx趋近于0〔1 - cos(2x)〕/(xsinx) -
范瑶19563289551…… 因为cos2x=cos²x-sin²x=1-2sin²x, 则有1-cos2x=2sin²x 所以(1-cos2x)/(xsinx)=(2sin²x)/(xsinx) =2sinx/x,根据两个重要极限,得出结果=2
@边叛5195：1 - cos2x极限怎么化解 -
范瑶19563289551…… 1-cos2x=2(sinx)^2 ,由此当X ->0时,其极限为0.........
@边叛5195：求函数lim x→0 1 - cos2x/xsinx 的极限
范瑶19563289551…… 等于2. 推导:cos2x=1-2*(sinx)^2 所以lim x→0 1-cos2x/xsinx =lim x→0 2*(sinx)^2/xsinx =2*sinx/x =2 因为lim x→0 sinx/x=1 你应该学过了.
范瑶19563289551…… lim(x趋向0)(1-cos2x)/xsinx =lim(x趋向0)[(1-1+2Sin^2(x)] /xsinx =lim(x趋向0)2sin^2x/xsinx =lim(x趋向0)2sinx/x =2
@边叛5195：lim(x→无穷)(1 - cos2x)/xsinx 有木有解,如何解? -
范瑶19563289551…… 有解.1-cos2x=2(sinx)^2, lim(x→无穷)(1-cos2x)/xsinx=lim(x→无穷)(2(sinx)^2)/xsinx=2lim(x→无穷)sinx/x, 因为|sinx|<=1,是有界变量,lim(x→无穷)1/x=0.是无穷小量,根据有界变量乘无穷小量仍为有界变量,得原式的极限为0
@边叛5195：x趋近于0 求(1 - cos2x)/(xsinx) 的极限 -
范瑶19563289551…… 解: lim【x→0】(1-cos2x)/(xsinx) =lim【x→0】2sin²x/(xsinx) =lim【x→0】(2sinx)/x =2 答案:2
@边叛5195：x趋近于0 求(1 - cos2x)/(xsinx) 的极限 - 作业帮
范瑶19563289551…… [答案] lim【x→0】(1-cos2x)/(xsinx) =lim【x→0】2sin²x/(xsinx) =lim【x→0】(2sinx)/x =2 答案:2
@边叛5195：求lim(x→∞)1 - cos2x/xsinx -
范瑶19563289551…… 1-cos2x=1-(1-2*(sinX)^2)=2*(sinX)^2,1-cos2x/xsinx=2sinx/x 因为当x→∞时,1/x→0 又sinx为有界函数,|sinx|≤1 所以lim【x→∞】sinx/x=0 lim(x→∞)1-cos2x/xsinx=0
@边叛5195：求极限limx 0时1 - cos2x/xsinx等于多少 -
范瑶19563289551…… 1-cos2x=2(sinx)^2利用等价无穷小式子=2x^2/x^2=2
@边叛5195：求lim(x→∞)1 - cos2x/xsinx - 作业帮
范瑶19563289551…… [答案] 1-cos2x=1-(1-2*(sinX)^2)=2*(sinX)^2,1-cos2x/xsinx=2sinx/x 因为当x→∞时,1/x→0 又sinx为有界函数,|sinx|≤1 所以lim【x→∞】sinx/x=0 lim(x→∞)1-cos2x/xsinx=0
@边叛5195：limx趋近于0〔1 - cos(2x)〕/(xsinx) -
范瑶19563289551…… 因为cos2x=cos²x-sin²x=1-2sin²x, 则有1-cos2x=2sin²x 所以(1-cos2x)/(xsinx)=(2sin²x)/(xsinx) =2sinx/x,根据两个重要极限,得出结果=2
@边叛5195：1 - cos2x极限怎么化解 -
范瑶19563289551…… 1-cos2x=2(sinx)^2 ,由此当X ->0时,其极限为0.........
@边叛5195：求函数lim x→0 1 - cos2x/xsinx 的极限
范瑶19563289551…… 等于2. 推导:cos2x=1-2*(sinx)^2 所以lim x→0 1-cos2x/xsinx =lim x→0 2*(sinx)^2/xsinx =2*sinx/x =2 因为lim x→0 sinx/x=1 你应该学过了.
