limarctanx趋于无穷

费施良3084x趋于1limarctanx+1 -
宿莺佩18522564670 ______ 无意中看到这道题.大家也有几种解法了: [(x+1)/(x+2)]-[1/(x+7)]=[(x+2)/(x+3)]-[1/(x+6)] 为简化起见,特设:(x+1)=a,(x+2)=b,(x+3)=c,(x+6)=d,(x+7)=e 于是公式成为:a/b-1/e=b/c-1/d 先移项:a/b-b/c=1/e-1/d 等式两边同乘以:bcde...

费施良3084求当x趋于0时,limarctan(1/x)=π/2的步骤 -
宿莺佩18522564670 ______ 这个极限是错误的 当x→0+时lim arctan(1/x)=π/2 当x→0-时lim arctan(1/x)=-π/2

费施良3084当x趋于0时,limarctan(1/x)=? -
宿莺佩18522564670 ______[答案] 当x趋于0时,lim arctan(1/x)不存在1.x -> 0+时,1/x -> +∞所以,arctan(1/x) -> π/2即lim(x->0+) arctan(1/x) = π/22.x -> 0-时,1/x -> -∞所以,arctan(1/x) -> -π/2即lim(x->0-) arctan(1/x) = -π/23.因为在x=0...

费施良3084求limarctan1/(x - 1)的极限,x趋于1+ -
宿莺佩18522564670 ______[答案] x→1+ x-1→0+ 所以1/(x-1)→+∞ 所以原式=π/2

费施良3084x趋近于0时limarctan1/x^2的极限 -
宿莺佩18522564670 ______ lim arctan1/x² = arctan(+∞) = ½π ± kπ (k= 0, 1, 2, 3, ....) x→0 通常在主值范围内考虑,是½π.

费施良3084求函数极限x→∞limarctanx/x谢了 -
宿莺佩18522564670 ______[答案] lim(x→∞)arctanx/x ∵arctanx为有界量,(1/x)趋于0 ∴lim(x→∞)arctanx/x=0

费施良3084limarctanx - x/ln(1+x^3)x趋于0 -
宿莺佩18522564670 ______[答案] lim(arctanx-x)/ln(1+x³) 【0/0,罗必塔法则】=lim[(1/1+x²)-1]/[3x²/(1+x³)]=lim(-x²/1+x²)/[3x²/(1+x³)]=lim(-1/3)*[(1+x³)/(1+x²)]=(-1/3)*[...

费施良3084arctanx和x为什么是等价无穷小
宿莺佩18522564670 ______ X→0时,arctanx~X令arctanx=y,x=tany,x趋于0时,y趋于0,因此limarctanx/x=limy/tany=limycosy/siny=limcosy/(siny/y)=1.即arctanx~x等价无穷小在求极限时有重要应用,定理如下:设在x的某一变化过程中,α和β都是无穷小,且α~α',β~β',存在(或为正无穷),则:lima/b=lima'/b'

费施良3084limx趋于负无穷acrtanx=? -
宿莺佩18522564670 ______ t-->-π/2 时 tan(t)->-∞ 即x=tan(t)->-∞ 所以arctanx=arctan(tan(t))->-π/2 limarctanx=-π/2

费施良3084当(x→0)时,证明 arctan x~x 求过程 -
宿莺佩18522564670 ______[答案] 令arctanx=u,则x=tanu,且当x→0,u→0 故 limarctanx/x=limu/tanu=limucosu/sinu=limu/sinu=1