利用
√(1+x) = 1 + x/2+ (x^2)(1/2)(-1/2)/2! + o(x^2),
e^x = 1 + x/1!+ (x^2)/2! + o(x^2),
cosx = 1 - (x^2)/2! +(x^4)/4! + o(x^6),
可得
1 + (x^2)/2 - √(1+x^2)
= 1 + (x^2)/2 - [1 + (x^2)/2+ (x^4)(1/2)(-1/2)/2! + o(x^4)]
= -(x^4)(1/8) + o(x^4),
cosx-e^(x^2)
= [1 - (x^2)/2! + o(x^4)] - [1 + (x^2)/1! + o(x^2)]
= -(3/2)(x^2)+o(x^2),
和等价无穷小
sinx ~ x (x→0)
的替换,可得
g.e. = lim(x→0)[1 + (x^2)/2 - √(1+x^2)]/{[cosx-e^(x^2)]*(x^2)}
= lim(x→0)[-(x^4)(1/8) + o(x^4)]/{[-(3/2)(x^2)+o(x^2)]*(x^2)}
= lim(x→0)[-(1/8) + o(x^4)/(x^4)]/[-(3/2)+o(x^2)/(x^2)]
= (1/8)/(3/2)
= ……
追问![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/54fbb2fb43166d22ac127327442309f79052d218?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
啊,知道了,谢谢哈
忘了公式了
追答 教材干啥用的?翻书啊!记
f(x) =√(1+x),
泰勒公式是啥样的?写一下就有:
f(x) = f(0) + [f'(0)/1!]x +[f"(0)/2!]x² + o(x²)
= ……