Question

一道高数题大神求解啊见图片

Answer 1

追问

答案是-3/2啊

这题放在了导数的题目里面是不是跟导数有什么联系呢？

追答

答案错了。

追问

好吧…

谢谢

Answer 2

lim(x->0) [ tan(tanx) - sin(sinx) ]/[ x( e^(-x^2) - 1) ]

x-> 0
tanx(tanx) ~ [x + (1/3)x^3 ] + (1/3)[x + (1/3)x^3 ]^3
~ x + (2/3)x^3
sin(sinx) ~ [x- (1/6)x^3] -(1/6)[x- (1/6)x^3]^3
~ x - (1/3)x^3

e^(-x^2) ~ 1-x^2

tan(tanx) - sin(sinx) ~ x^3
x( e^(-x^2) -1) ~ x^3

lim(x->0) [ tan(tanx) - sin(sinx) ]/[ x( e^(-x^2) - 1) ]
=lim(x->0) x^3/x^3
=1

追问

那个tan(tanx)和sin(sinx)的变换是怎么来的啊？

还有答案上说这是-3/2

追答

lim(x->0) [ tan(tanx) - sin(sinx) ]/[ x( e^(-x^2) - 1) ]

x-> 0
tanx(tanx) ~ [x + (1/3)x^3 ] + (1/3)[x + (1/3)x^3 ]^3
~ x + (2/3)x^3
sin(sinx) ~ [x- (1/6)x^3] -(1/6)[x- (1/6)x^3]^3
~ x - (1/3)x^3

e^(-x^2) ~ 1-x^2

tan(tanx) - sin(sinx) ~ x^3
x( e^(-x^2) -1) ~ -x^3

lim(x->0) [ tan(tanx) - sin(sinx) ]/[ x( e^(-x^2) - 1) ]
=lim(x->0) x^3/(-x^3)
=-1

e.g
tanx ~ x + (1/3)x^3
tan(tanx) ~[x + (1/3)x^3] + (1/3)[x + (1/3)x^3]^3
~ x+ (2/3)x^3
sinx ~ x - (1/6)x^3
sin(sinx) ~[x - (1/6)x^3] - (1/6)[x - (1/6)x^3]^3
~ x - (1/3)x^3

追问

懂了谢谢