Question

求极限题目求解

求极限题目求解求这三道题的完整解题过程!

Answer 1

分母的极限是0，分子的极限是ln tan1.而tan1近似等于1.55，再对1.55取自然对数时分子的极限肯定不等于0。所以不满足洛比达法则。极限应该是无穷大。

Answer 2

1.
原式 = lim(x->0) 2(sinx-x*cosx)/(x^2 *sin2x)
= lim(x->0) 2(cosx-cosx+x*sinx)/(2x*sin2x +x^2 *2cos2x)
= lim(x->0) sinx/(sin2x+x*cos2x)
= lim(x->0) cosx/(2cos2x+cos2x-2x*sin2x)
=1/3
2.
原式 = lim(x->1) (3x^2 -3)/(3x^2 -2x-1)
= lim(x->1) 6x/(6x-2)
=3/2
3.
原式 = lim(x->0) (sinx-e^x +1)/[sinx*(e^x -1)]
= lim(x->0) (cosx-e^x)/[cosx*(e^x -1)+sinx*e^x]
= lim(x->0) (-sinx-e^x)/[-sinx*(e^x -1)+2cosx*e^x+sinx*e^x]
=-1/2

Answer 3

(1)
lim(x->0) (tanx-x) /(x^2.sinx)
=lim(x->0) (tanx-x) /x^3 (0/0)
=lim(x->0) [(secx)^2-1 ] /(3x^2)
=lim(x->0) (tanx)^2 /(3x^2)
=lim(x->0) x^2 /(3x^2)
=1/3
(2)
lim(x->1) (x^3-3x+2)/(x^3-x^2-x+1)
=lim(x->1) (x-1)^2.(x+2)/[ (x-1)^2.(x+1)]
=lim(x->1) (x+2)/(x+1)
=3/2
(3)
lim(x->0) [1/(e^x-1) -1/sinx ]
=lim(x->0) (sinx-e^x+1 )/[ sinx.(e^x-1)]
=lim(x->0) (sinx-e^x+1 )/ x^2 (0/0)
=lim(x->0) (cosx-e^x )/ (2x) (0/0)
=lim(x->0) (-sinx-e^x )/ 2
=-1/2