Question

高数求极限？

题目如图

Answer 1

该极限可以第一步，提出sinx
原式=sinx*[x^2-ln2(1+x)]/x^4
=[x^2-ln2(1+x)]/x^3
洛必达法则求导
=[2x-2ln(1+x)/(1+x)]/3/x^2

Answer 2

原式=lim(x->0) sinx*[x^2-ln^2(1+x)]/[x^2*ln^2(1+x)]
=lim(x->0) [x^2-ln^2(1+x)]/[x*ln^2(x+1)]
=lim(x->0) [2x-2ln(1+x)/(1+x)]/[ln^2(x+1)+2xln(x+1)/(x+1)]
=2*lim(x->0) [x+x^2-ln(1+x)]/[(x+1)ln^2(x+1)+2xln(x+1)]
=2*lim(x->0) [1+2x-1/(1+x)]/[ln^2(x+1)+4ln(x+1)+2x/(x+1)]
=2*lim(x->0) (3x+2x^2)/[(x+1)ln^2(x+1)+4(x+1)ln(x+1)+2x]
=2*lim(x->0) (3+4x)/[ln^2(x+1)+6ln(x+1)+6]
=2*(3/6)
=1