Question

大一高数，求大神

Answer 1

x->0
分子
tanx = x+(1/3)x^3+o(x^3)
sinx = x-(1/6)x^3+o(x^3)
tanx - sinx = (1/2)x^3 +o(x^3)
分母
ln(1+x) = x -(1/2)x^2 +o(x^2)
xln(1+x) =x^2 -(1/2)x^3 +o(x^3)
xln(1+x) -x^2= -(1/2)x^3 +o(x^3)
lim(x->0) [√(1+tanx) -√(1+sinx) ]/[xln(1+x) -x^2]
=lim(x->0) [(1+tanx) -(1+sinx) ]/ { [xln(1+x) -x^2] .[√(1+tanx) +√(1+sinx) ] }
=(1/2) lim(x->0) [(1+tanx) -(1+sinx) ]/ [xln(1+x) -x^2]
=(1/2) lim(x->0) (tanx-sinx)/ [xln(1+x) -x^2]
=(1/2) lim(x->0) (1/2)x^3/ [(-1/2)x^3 ]
=-1/2