一道高数求极限的题目ln(sin눀x+e^x)-x⼀x눀+e^2x-2x

加减的时候不宜等价替换

x->0
分子
(sinx)^2 = x^2 +o(x^2)
e^x = 1+ x +(1/2)x^2 +o(x^2)
(sinx)^2 + e^x = 1+x + (3/2)x^2 +o(x^2)
ln[ (sinx)^2 +e^x ]
=ln[1+x + (3/2)x^2 +o(x^2)]
=[x + (3/2)x^2 ] -(1/2)[x + (3/2)x^2 ]^2 +o(x^2)
=[x + (3/2)x^2 ] -(1/2)[x^2 +o(x^2) ] +o(x^2)
= x + x^2 +o(x^2)
ln[ (sinx)^2 +e^x ] -x =x^2 +o(x^2)
分母

e^(2x) = 1+2x+ 2x^2 +o(x^2)
x^2+e^(2x) =1+2x+ 3x^2 +o(x^2)
ln[x^2+e^(2x)]
=ln[1+2x+ 3x^2 +o(x^2)]
=(2x+ 3x^2) -(1/2)(2x+ 3x^2)^2 +o(x^2)
=(2x+ 3x^2) -(1/2)[4x^2+o(x^2) ] +o(x^2)
=2x +x^2 +o(x^2)
ln[x^2+e^(2x)] -2x =x^2 +o(x^2)
lim(x->0) { ln[ (sinx)^2 +e^x ] -x } / { ln[x^2+e^(2x)] -2x }
=lim(x->0) x^2 / x^2
=1

你用到了等价无穷小ln（1+x）～x，但是等价有前提，只有乘除的情况才可以用，后面有-2x，不可以直接用等价无穷小