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求极限lim(x→0)(sinx%xCosx)/(sin^3x),答案是1/3

lim(x→0)(sinx-xcosx)/(sin^3x) =lim(x→0)[(sinx-xcosx)]'/(sin^3x)' =lim(x→0)(cosx-cosx+xsinx)/[3(sin^2x)cosx] =lim(x-0) x/[3(sinxcosx)] =lim(x-0) x/[3sin(2x)/2] =lim(x-0) x'/[3sin(2x)/2]' =lim(x-0) 1/[3cos(2x)] =1/3

limx→0 (sinx-xcosx)/sin^3x =(1-xcotx)/sin²x =(tanx-x)/x³ 利用等价无穷小:sinx∽x∽tanx =(sec²x-1)/3x² 洛必达法则,上下求导 =tan²x/3x² =1/3 利用等价无穷小:x∽tanx

lim(x→0)(sinx-xcosx)/(sin^3x)=lim(x→0)[(sinx-xcosx)]'/(sin^3x)'=lim(x→0)(cosx-cosx+xsinx)/[3(sin^2x)cosx]=lim(x-0) x/[3(sinxcosx)]=lim(x-0) x/[3sin(2x)/2]=lim(x-0) x'/[3sin(2x)/2]'=lim(x-0) 1/[3cos(2x)]=1/3

为什么会等于0?你从哪里得到的?加减法中等价替换?

(x→0)lim [sin²x/x] = (x→0)lim [x²/x] = (x→0)lim [x] = 0 罗必塔: (x→0)lim [sin²x/x] = (x→0)lim [2sinxcosx/1] = (x→0)lim [sin2x] = sin0 = 0

第一步,(1-cosxcos2xcos3x一直乘到cosnx)和二分之一倍的(1平方 加 2平方 加加加加到 n平方)倍的x平方 是等价无穷小,具体的证明你可以用 ln(1加x)和x 这对等价无穷小(PS:这么代换有很大的好处,大家去试一下就明白了) 去证,,,不明白的...

lim(x→0+) ln(sin3x)/ln(sinx) =lim(x→0+) [3cos3x/(sin3x)/[cosx/sinx] =lim(x→0+) (3sinx/sin3x=1

limx趋向于0。1/(sin方x)-1/(x方乘以cos方x) =lim [(sin方x)-(x方乘以cos方x)]/[(sin方x)(x方乘以cos方x)] =lim [sin²x-x²(1-sin²x)]/[sin²x*x²(1-sin²x)] =lim [sin²x-x²+x²sin²x)]/[sin...

先看第一步tanx-sinx就是公式变形,sinx=tanx*cosx,然后代进去,tanx-tanx*cosx tanx(1-cosx),然后tanx等价于x,1-cosx等价于 2x^2, sin^3x 等价于x^3后,当X->0时,lim(x->0)(x*2x^2)\x^3=1\2

请看图

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