求极限 lim[x-x^2ln(1+(1/x))] x->∞
答案:1 悬赏:50 手机版
解决时间 2021-03-25 00:59
- 提问者网友：趣果有间
- 2021-03-24 17:32
求极限 lim[x-x^2ln(1+(1/x))] x->∞
最佳答案
- 五星知识达人网友：低音帝王
- 2021-03-24 18:22
lim_{x->∞}[x-x^2ln(1+(1/x))]
=lim_{x->∞}[1/x-ln(1+(1/x))]/(1/x^2)
=lim_{t->0}[t-ln(1+t)]/(t^2)
=lim_{t->0}[1-1/(1+t)]/(2t)
=lim_{t->0}[1/(1+t)]/2
= 1/2
lim_{x->0}(1/sin^2(x)-1/x^2)
= lim_{x->0}[(x^2 - (sinx)^2)/(xsinx)^2]
= lim_{x->0}[(x^2 - (sinx)^2)/x^4]
= lim_{x->0}[(2x - 2(sinx)(cosx))/(4x^3)]
= lim_{x->0}[(2x - sin(2x))/(4x^3)]
= lim_{x->0}[(2 - 2cos(2x))/(12x^2)]
= lim_{x->0}[(1 - cos(2x))/(6x^2)]
= lim_{x->0}[2sin(2x)/(12x)]
= lim_{x->0}[sin(2x)/(2x)]/3
= 1/3.
(cotx)^(1/lnx) = exp{ln(cotx)/lnx}
lim_{x->0+}ln(cotx)/lnx
= lim_{x->0+}[ln(cosx) - ln(sinx)]/lnx
= lim_{x->0+}[-sinx/cosx - cosx/sinx]/(1/x)
= lim_{x->0+}[-xsinx/cosx - xcosx/sinx]
= 0 - 1
= -1.
lim_{x->0+}(cotx)^(1/lnx)
= exp{lim_{x->0+}ln(cotx)/lnx }
= e^(-1).
=lim_{x->∞}[1/x-ln(1+(1/x))]/(1/x^2)
=lim_{t->0}[t-ln(1+t)]/(t^2)
=lim_{t->0}[1-1/(1+t)]/(2t)
=lim_{t->0}[1/(1+t)]/2
= 1/2
lim_{x->0}(1/sin^2(x)-1/x^2)
= lim_{x->0}[(x^2 - (sinx)^2)/(xsinx)^2]
= lim_{x->0}[(x^2 - (sinx)^2)/x^4]
= lim_{x->0}[(2x - 2(sinx)(cosx))/(4x^3)]
= lim_{x->0}[(2x - sin(2x))/(4x^3)]
= lim_{x->0}[(2 - 2cos(2x))/(12x^2)]
= lim_{x->0}[(1 - cos(2x))/(6x^2)]
= lim_{x->0}[2sin(2x)/(12x)]
= lim_{x->0}[sin(2x)/(2x)]/3
= 1/3.
(cotx)^(1/lnx) = exp{ln(cotx)/lnx}
lim_{x->0+}ln(cotx)/lnx
= lim_{x->0+}[ln(cosx) - ln(sinx)]/lnx
= lim_{x->0+}[-sinx/cosx - cosx/sinx]/(1/x)
= lim_{x->0+}[-xsinx/cosx - xcosx/sinx]
= 0 - 1
= -1.
lim_{x->0+}(cotx)^(1/lnx)
= exp{lim_{x->0+}ln(cotx)/lnx }
= e^(-1).
我要举报
如以上问答信息为低俗、色情、不良、暴力、侵权、涉及违法等信息，可以点下面链接进行举报！
大家都在看
推荐资讯