用洛必达法则求极限,打勾的

2025-05-07 10:08:47
推荐回答(2个)
回答1:

①=4

回答2:

(1) 由罗必塔法则得 原式 = lim[1-1/(1+x^2)]/[1-1/√(1-x^2)]
= limx^2√(1-x^2)/{[√(1-x^2)-1](1+x^2)}
= limx^2/[√(1-x^2)-1] = limx^2[√(1-x^2)+1]/(-x^2)
= - lim√(1-x^2)+1 = -2.
(3) 原式 = lim(x-arcsinx)/x^3 由罗必塔法则得
= lim[1-1/√(1-x^2)]/(3x^2) = lim[√(1-x^2)-1]/[3x^2√(1-x^2)]
= lim[√(1-x^2)-1]/(3x^2) = lim-x^2/{3x^2[√(1-x^2)+1]}
= lim-1/{3[√(1-x^2)+1]} = -1/6.
(4) 原式 = limx^2/(secx-cosx) = limx^2 cosx/[1-(cosx)^2]
= limx^2/(sinx)^2 = 1 , 不用罗必塔法则更简单。