Question

大学高数极限题，3题

Answer 1

（1）因为x-lnx<x-lnx*sinx<x+lnx
且lim(x->+∞) x-lnx
=lim(x->+∞) x(1-lnx/x)
=+∞
lim(x->+∞) x+lnx=+∞
所以根据夹逼性，lim(x->+∞) x-lnx*sinx=+∞
原式=π/2
（3）原式=lim(x->0) [x^2/2+1-1-x^2/2+x^4/8+o(x^4)]/[1-x^2/2+o(x^3)-1-x^2+o(x^2)]x^2
=lim(x->0) [x^4/8+o(x^4)]/[-(3/2)*x^2+o(x^2)]x^2
=lim(x->0) [x^2/8+o(x^2)]/[-(3/2)*x^2+o(x^2)]
=-1/12
（5）令t=π/2-x，则x=π/2-t
原式=lim(t->0) [lnsin(π/2-t)]/4t^2
=lim(t->0) [lncost]/4t^2
=lim(t->0) -tant/8t
=lim(t->0) -t/8t
=-1/8

Answer 2