limx趋近于1 sin（2x-1）
17088368500&&已知lim x趋近于1sin2(x - 1)╱x2 ax b=1求a.b - 》》》 lim(x->1)sin2(x-1)/(x^2 ax b) (0/0) => 1 a b=0 a b=-1 (1) lim(x->1)sin2(x-1)/(x^2 ax b) (0/0) =lim(x->1)2cos2(x-1)/(2x a) =2/(2 a) = 1 => a= 0 from (1) a b=-1 b=-1
17088368500&&limx趋近于1,sin^2(x—1)/x^2—1=? - 》》》 洛比达法则,上下求导,再将1代入
17088368500&&lim(x趋近于0)((1/sin^2 x) - (cos^2 x)/x^2)= - 》》》 lim(x→0)((1/sin^2 x)-(cos^2 x)/x^2)=lim(x→0) ((x^2-sin^2xcos^2x)/x^2*sin^2x)=lim(x→0)((x^2-1/4sin^2(2x))/x^4) sinx与x等价无穷小量,再用罗必塔=lim(x→0)(2x-1/2sin4x)/4x^3=lim(x→0)(2-2cos4x)/12x^2=lim(x→0)(8sin4x/24x)=lim(x→0)(32cos4x/24)=32/24=4/3
17088368500&&函数极限:sin2(x - 1)/x - 1的x趋向于1的极限 sin后面的2是平方 - 》》》 t=x-1 sin2t/t=2sintcost/t t==>0,c0st==>1 sint/t ==1 极限为2
17088368500&&lim(x趋向1)sin(x^2 - 1)/tan(x - 1) 求洛必达法则的解法与不使用洛必达法则的解法 - 》》》[答案] (1) lim(x趋向1)sin(x²-1)/tan(x-1) =lim(x趋向1)2x•cos(x²-1)/[1/cos²(x-1)] =lim(x趋向1)2x•cos(x²-1)cos²(x-1)=2 (2) x→1时,sin(x²-1)~x²-1,tan(x-1)~x-1 所以 lim(x趋向1)sin(x^2-1)/tan(x-1)=lim(x趋向1)( x²-1)/(x-1)=lim(x趋向1)( x 1)=2
17088368500&&limx趋近于无穷(1/x)sin((x^2 1)/x)要过程 - 》》》 因为sin((x^2 1)/x)的绝对值小于等于1,所以 limx趋近于无穷(1/x)sin((x^2 1)/x)的绝对值≤limx趋近于无穷(1/x)的绝对值=0,所以,limx趋近于无穷(1/x)sin((x^2 1)/x)=0
17088368500&&limx趋于1 sin兀x/1 - x^2 - 》》》[答案] 令x-1=t x=t 1 原式=lim(t->0) sinπ(t 1)/[1-(t 1)^2] =lim(t->0) -sinπt/[-2t-t^2] =lim(t->0) (πt)/(2t) =π/2
17088368500&&limx趋向于0(1/sin^(2)x - 1/x^2) 洛必达法则 - 》》》[答案] lim(x->0)(1/(sinx)^2-1/x^2) =lim(x->0)[x^2-(sinx)^2]/x^4 =lim(x->0)(x sinx)/x*lim(x->0)(x-sinx)/x^3 =2lim(x->0)(1-cosx)/(3x^2) =2/3*lim(x->0)1/2x^2/x^2 =1/3
17088368500&&limx趋近于0,sin2xcosx/cos2xsinx的极限 - 》》》 原式=lim(x→0) (2sinxcosx*cosx)/(cos2x*sinx) =lim(x→0) 2(cosx)^2/cos2x =2*1/1 =2 望采纳
17088368500&&当x趋于1时lim1/x–1sin(2x–2)的极限 - 》》》[答案] 原式=lim(x->1)sin2(x-1)/(x-1) =lim(x->1)sin2(x-1)/2(x-1) *2 =2*1 =2