1-+cos+2+x等价于多少

索房汤964limx趋于0+的1 - cos2x/根号下1+cosx -
秦青贺19537411339 ______ 显然x趋于 那么√[1+f(x)(1-cos2x)] - 1也趋于0, 而且等价于0.5* f(x)(1-cos2x), 而1-cos2x等价于0.5*(2x)^2即2x^2, 所以√[1+f(x)(1-cos2x)] - 1等价于 x^2 *f(x), 而x趋于0时,sinx也是x的等价无穷小, 故 lim(x趋于0) {√[1+f(x)(1-cos2x)] - 1} /xsinx =lim(x趋于0) x^2 *f(x) /x^2 =lim(x趋于0) f(x) =6

索房汤9641 - cosx∼x²/2,那么1 - cos2x等价谁 -
秦青贺19537411339 ______[答案] (2x)²/2=2x² 前提是x→0

索房汤964lim在X趋于1时,利用等价无穷小的性质求(1+cos兀x)/(x - 1)^2 -
秦青贺19537411339 ______ x→1 lim (1+cosπx) / (x-1)^2 换元,t=1-x,x=1-t =lim(t→1) (1+cosπ(1-t)) / t^2 =lim (1-cosπt) / t^2 利用等价无穷小:1-cosx~x^2/2 =lim (πt)^2 / 2t^2 =lim π^2/2 =π^2/2 有不懂欢迎追问

索房汤964lim(x趋向0)[1 - cos(1 - cos2x)]/x^4怎么解? -
秦青贺19537411339 ______[答案] 用等价无穷小lim(x趋向0)[1-cosx]等价于lim(x趋向0)[(x^2)/2]lim(x趋向0)[1-cos(1-cos2x)]/x^4=lim(x趋向0)[1-cos(2x^2)]/x^4=lim(x趋向0)[1-cos(2x^2)]/x^4=lim(x趋向0)[(4x^4)/x^4]=4

索房汤964利用等价无穷小的代换性质求,当x趋于1时,(1+cosπx)/(x - 1)^2的极限是多少..... -
秦青贺19537411339 ______[答案] 令x-1=t,则 当x趋于1时,t趋于0 (1+cosπx)/(x-1)^2 =(1+cosπ(t+1))/t^2 =(1-cost)/t^2 1-cost等价于t^2/2 所以 原式的极限=1/2.

索房汤964limx趋近于0〔1 - cos(2x)〕/(xsinx) -
秦青贺19537411339 ______ 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

索房汤964如何证明x趋向于0的时候1 - cosx与1/2x^2是等价无穷小. -
秦青贺19537411339 ______ 证明:cosx=cos(x/2+x/2)=cos(x/2)^2-sin(x/2)^2=2cos(x/2)^2-1所以,左式=2-2cos(x/2)^2=2sin(x/2)^2x趋向于0时,上式=2*(x/2)^2=1/2 x^2得证

索房汤964高数等价无穷小(1+cosx)ln(1+x)和谁等价怎么算
秦青贺19537411339 ______ 在两种情况下 ① x→0;② x→(2n+1)π,n≥0.f(x)=(1+cosx)ln(1+x) 是无穷小. ①lim[(1+cosx)ln(1+x)]/x=2, 当x→0时,(1+cosx)ln(1+x)~2x; ②lim[(1+cosx)ln(1+x)]/[x-(2n+1)π]^2=ln[1+(2n+1)π]/2, 当x→(2n+1)π时(n≥0),(1+cosx)ln(1+x)~{ln[1+(2n+1)π]/2}[x-(2n+1)π]^2. .

索房汤964lim(x→0)(1 - cos2x)/(x*sin2x) lim(x→0)(xcot2x) lim(x→无穷)(1+(1/2x))^x -
秦青贺19537411339 ______ lim(x→0)(1-cos2x)/(x*sin2x)=lim(x→0)2sin²x/(x*sin2x) 用等价无穷小=lim(x→0)2x²/(x*2x)=1 lim(x→0)(xcot2x)=lim(x→0)x/tan2x 用等价无穷小=lim(x→0)x/2x=1/2 lim(x→∞)(1+(1/2x))^x=lim(x→∞)(1+(1/2x))^(2x/2) 用重要极限lim(x→∞)(1+1/x)^x=e=e^(1/2)=√e

索房汤9641 - cos(ln(1+X^2))是X的( )阶无穷小 求详细解答过程 -
秦青贺19537411339 ______ 1-cos(ln(1+X^2)) 等价于【ln(1+x²)】²/2等价于x的4次方/2所以1-cos(ln(1+X^2))是X的( 4)阶无穷小