ruoshi14
幼苗
共回答了13个问题采纳率:76.9% 举报
第1题是不对的
因为lim 1+cos2x+sin3x-1=1
lim [cot5x/(cos2x+sin3x-1)](cos2x+sin3x-1)→∞
所以是1^∞型极限,运用重要极限 lim(t→0)(1+t)^(1/t)=e:
lim(1+cos2x+sin3x-1)^{[cot5x/(cos2x+sin3x-1)](cos2x+sin3x-1)}
=lim(1+cos2x+sin3x-1)^{[1/(cos2x+sin3x-1)]*[cot5x(cos2x+sin3x-1)]}
注意到lim(1+cos2x+sin3x-1)^{[1/(cos2x+sin3x-1)]=e (重要极限中t=cos2x+sin3x-1)
所以
lim(1+cos2x+sin3x-1)^{[cot5x/(cos2x+sin3x-1)](cos2x+sin3x-1)}
=lim(1+cos2x+sin3x-1)^{[1/(cos2x+sin3x-1)]*[cot5x(cos2x+sin3x-1)]}
=e^lim [cot5x(cos2x+sin3x-1)]
=e^lim [(cos2x+sin3x-1)/sin5x]cos5x
=e^lim (-2sin2x+3cos3x)/5sec²5x
=e^(3/5)
第2题是用的等价无穷小替换
x→0时,ln(1+x²)~x²
∴lim ln(1+x^2)/y(x)=lim x^2/y(x)
1年前
4