lim(x→0)e^sinx-e^tanx⼀x(1-cos4x)

lim(x→0)e^sinx-e^tanx/x(1-cos4x)
2024-11-08 15:01:28
推荐回答(2个)
回答(1):

简单计算一下即可,答案如图所示

回答(2):

lim(x→0)[e^sinx-e^tanx]/[x(1-cos4x)]

x->0
sinx ~ x -(1/6)x^3
e^sinx ~ 1 + (x -(1/6)x^3) + (x -(1/6)x^3)^2/4
~ 1+x +x^2/4 -(1/6)x^3

tanx ~ x + (1/3)x^3
e^tanx ~1 + (x +(1/3)x^3) + (x +(1/3)x^3)^2/4
~ 1+x +x^2/4 +(1/3)x^3

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

1-cos4x ~ 1 -( 1 -(1/2)(4x)^2 )
= 8x^2
x(1-cos4x) ~ 8x^3

lim(x→0)[e^sinx-e^tanx]/[x(1-cos4x)]
=lim(x→0)[-(1/2)x^3/[8x^3]
=-1/16