当x趋向于0时,求:(1\/sin^2x-1\/x^2)的极限?
分子要用无穷小代换的话,必须对正弦函数展开式展开到与分母同阶,舍弃更高阶的无穷小才可以,否则会发生错误。
当x趋近于0时,求(1\/sin^2x)-1\/x^2的极限
x^2*sin^2x等价于x^4 所以原极限=lim(x趋于0) 2x *(x -sinx) \/ x^4 =lim(x趋于0) 2(x -sinx) \/ x^3 所以洛必达法则求导 =lim(x趋于0) 2(1 -cosx) \/ 3x^2 而1-cosx等价于0.5x^2,代入得到原极限= 2* 0.5x^2 \/3x^2= 1\/3 ...
当x趋向于0,求1\/sin^2-1\/x^2的极限.洛必达求导不来.
lim(1\/sin^2x-1\/x^2)=lim(x^2-sin^2x)\/(x^2sin^2x)=lim(x^2-sin^2x)\/(x^4)=lim(2x-sin2x)\/4x^3 =lim(2-2cos2x)\/12x^2 =lim4sin2x\/24x =1\/3
1\/sin^2x-1\/x^2 x趋向0求极限
lim(1\/sin^2x-1\/x^2)=lim(x^2-sin^2x)\/x^4 =lim(2x-2sinxcosx)\/4x^3 =lim(2x-sin2x)\/4x^3 =lim(2-2cos2x)\/12x^2 =lim4sin2x\/24x =4\/12=1\/3
lim(X→0)(1\/sin^2X-1\/X^2)
lim[x→0] (1\/sin²x - 1\/x²)=lim[x→0] (x²-sin²x)\/(x²sin²x)分母等价无穷小代换 =lim[x→0] (x²-sin²x)\/x⁴洛必达法则 =lim[x→0] (2x-2sinxcosx)\/(4x³)=lim[x→0] (2x-sin2x)\/(4x³)洛必...
洛必达定理当x趋于0时,求1\/(sinx)^2-1\/x^2的极限求详解
原式=(x²-sin²x)\/x²sin²x 0\/0型 分子求导=2x-2sinxcosx=2x-sin2x 分母求导=2xsin²x+x²sin2x 还是0\/0型 分子求导=2-2cos2x 分母求导=2sin²x+4xsin2x+2x²cos2x 还是0\/0型 分子求导=4sin2x 分母求导=6sin2x+12xcos2x-4x...
(1\/(sinx)^2-1\/(x^2))当X趋于0时的极限是多少最好有过程?非常感谢...
x->0 lim[1\/(sinx)^2-1\/x^2]=lim[(x^2-(sinx)^2)\/(x^2(sinx)^2)]=lim[(x^2-(sinx)^2)\/x^4]lim[x^2\/(sinx)^2]=lim[(x^2-(sinx)^2)\/x^4](之后的使用洛比达法则)=lim[(2x-2sinxcosx)\/(4x^3)]=lim[(2-2cos2x)\/(12x^2)]=lim[(4sin2x)\/(24x)]=(1...
当x趋于零时, 求sin平方分之一 减去 x平方分之一 的极限..
解:(x→0)时,原极限=1\/(sinx)^2-1\/x^2=[x^2-(sinx)^2]\/x^4 =(2x-sin2x)\/(4x^3)=(2-2cos2x)\/12x^2 (两次运用洛必达法则)=2*(2x^2)\/(12x^2) {(x→0)时,1-cos2x等价于(2x)^2\/2=2x^2} =1\/3 求极限基本方法有:1、分式中,分子分母同除以最高次,化无...
用洛必塔法则 求极限 lim(x趋于0) (1\/sinx^2)-(1\/x^2) 求解步骤
=lim(x→0) (x^2-sinx^2)\/(x^2sin^2x) (等价无穷小代换)=lim(x→0) (x^2-sinx^2)\/(x^4) (0\/0,洛必达法则)=lim(x→0) (2x-2sinxcosx)\/(4x^3)=lim(x→0) (x-1\/2sin2x)\/(2x^3) (0\/0,洛必达法则)=lim(x→0) (1-cos2x)\/(6x^2) (...
用洛必塔法则 求极限 lim(x趋于0) (1\/sinx^2)-(1\/x^2)
=lim(x→0) (x^2-sinx^2)\/(x^2sin^2x) (等价无穷小代换)=lim(x→0) (x^2-sinx^2)\/(x^4) (0\/0,洛必达法则)=lim(x→0) (2x-2sinxcosx)\/(4x^3)=lim(x→0) (x-1\/2sin2x)\/(2x^3) (0\/0,洛必达法则)=lim(x→0) (1-cos2x)\/(6x^2) (等价无穷小代换...