求极限lim[x-x^2ln(1+(1/x))]x->∞还有2个题:lim(1/sin^2(x)-1/x^2)x->0lim(cotx)^(1/lnx)x->0+答案分别为：1/2,1/3,e^(-1)

Question

求极限 lim[x-x^2ln(1+(1/x))] x->∞
还有2个题:lim(1/sin^2(x)-1/x^2) x->0
lim(cotx)^(1/lnx) x->0+
答案分别为：1/2,1/3 ,e^(-1)

Answer 1

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).