求解1.lim x→-2 (tanπx)/(x+2) 2.lim x→0 cot2xcot[(π/2)-x] 3.lim x→π/3 (1-2cosx)/(π-3x)
作者&投稿:叱干丹 (若有异议请与网页底部的电邮联系)
lim(x→0)(1/x^2-(cotx)^2)
=lim(x→0)[(cotx)^2-x^2]/x^2(cotx)^2
=lim(x→0)[(cosx)^2-x^2sinx^2]/[x^2cosx^2]
=lim(x→0)[-x^2+(1+x^2)cosx^2]/[x^2cosx^2]
=+∞
lim(x→1) (1-x)(tan(πx/2)
=lim(x→1)(1-x)/cot(πx/2)
=lim(x→1)(1-x)'/cot(πx/2)'
=lim(x→1) (-1)/[(π/2)/-(sinπx/2)^2]
=2/π
lim(x->1) (1-x) tan(πx/2)
=lim(x->1) (1-x) / cot(πx/2) (0/0)
=lim(x->1) -1 /(-[csc(πx/2)]^2 )
=lim(x->1) [sin(πx/2)]^2
=1
2.=tanx/tan2x=x/2x=1/2
3.洛比达法则 2sinx/-3=-根号3/3
欢迎追问!
兆德猗清: tan(πx/2)=cot(π/2-πx/2)=1/tan(π/2-πx/2)等价于1/(π/2-πx/2) ∴原式=lim(x→1)(1-x)/(π/2-πx/2)=2/π
娄烦县17582168653: 求解1.lim x→ - 2 (tanπx)/(x+2) 2.lim x→0 cot2xcot[(π/2) - x] 3.lim x→π/3 (1 - 2cosx)/(π - 3x) - ?
兆德猗清: 1.洛比达法则 π/cos^2πx=π2.=tanx/tan2x=x/2x=1/23.洛比达法则 2sinx/-3=-根号3/3 欢迎追问!
娄烦县17582168653: lim x→ - 2 (tanπx)/(x+2) 2.lim x→0 cot2xcot[(π/2) - x] 3.lim x→π/3 (1 - 2cosx)/(π - 3x) - ?
兆德猗清:[答案] 1.洛比达法则 π/cos^2πx=π 2.=tanx/tan2x=x/2x=1/2 3.洛比达法则 2sinx/-3=-根号3/3 欢迎追问!
娄烦县17582168653: lim x→1 (1 - x)tan[(πx)/2]用洛必达法则如何求解 - ?
兆德猗清: =(1-x)/cot(πx/2)=-1/[-csc²(πx/2)*(π/2)] =2/π
娄烦县17582168653: 求lim(1 - x)tanπx/2 x→1 - ?
兆德猗清: 解:lim(x→1)(1-x)tan(πx/2) =lim(x→1)(1-x)/cot(πx/2) =2/πlim(x→1)(-1)/-csc^2(πx/2) =2/π
娄烦县17582168653: 求x→1时lim(2 - x)^tan(πx)/2的极限(能不能不用若必达法则),谢谢 - ?
兆德猗清: 先说明(1):x→1时lim(1-x)/sin[π/2(1-x)]=lim[π/2(1-x)]/π/2sin[π/2(1-x)]=2/π 重要极限公式x→0时limsinx/x=1所得, (2)tan(πx)/2=[sin(πx)/2]/[cos(πx)/2]=[sin(πx)/2]/sin[π/2-(πx)/2] =[sin(πx)/2]/sin[π/2(1-x)] 所以x→1时lim(1-x)/tan(πx)/2=lim(1-x...
娄烦县17582168653: 用洛必达法则求下列极限(1)lim(x→0^+)lntan7x/lntan2x(2)lim(x→0)xcot2x - ?
兆德猗清:[答案] (1)lim(x→0^+)lntan7x/lntan2x=lim [7(sec7x)^2 / tan7x ] / [2(sec2x)^2 / tan2x ] =lim [7(sec7x)^2 *tan2x] / [2(sec2x)^2* tan7x ] 罗比达法则=lim [7(sec7x)^2 *2x] / [2(sec2x)^2* 7x ] 等价替换lim [7 *2] ...
娄烦县17582168653: 求极限:lim(1 - x)tan Π/2 - ?
兆德猗清:[答案] tan(πx/2)吧? lim(x→1) (1-x)tan(πx/2)=lim(x→1) (1-x) sin(πx/2)/cos(πx/2)=lim(x→1) (1-x) /cos(πx/2)=lim(x→1) (1-x) /sin(π/2-πx/2)=lim(x→1) (1-x) /(π/2-πx/2)=2/π希望采纳...
娄烦县17582168653: 求lim(1 - x)tanπx/2 x→1 - ?
兆德猗清:[答案] lim(x→1)(1-x)tan(πx/2) =lim(x→1)(1-x)/cot(πx/2) =2/πlim(x→1)(-1)/-csc^2(πx/2) =2/π
娄烦县17582168653: 当x趋向1时,求极限lim(1 - x)tan(πx/2),求详细过程~ - ?
兆德猗清:[答案] 原式=lim(1-x)sin(πx/2)/cos(πx/2) 是0/0型,用洛必达法则 =lim[-sin(πx/2)+(1-x)πcos(πx/2)/2]/[-πsin(πx/2)/2] =1/(π/2) =2/π