Question

洛必达法则，高等数学

求这几道题的解答过程

Answer 1

看图！

Answer 2

(x→π/4)lim(tanx)^tan2x=(x→π/4)lim(1+tanx-1)^tan2x
由公式,当f(x)→0,g(x)→∞时 lim [1+f(x)]＾g(x)=e＾lim(g(x)*ln[1+f(x)]) =e＾lim[g(x)*f(x)] 
所以原式＝e＾(x→π/4)lim[tan2x(tanx-1)]=e＾(x→π/4)lim[（2tanx/(1-tan²x)）*（tanx-1)]
＝e＾(x→π/4)lim[(-2tanx/(1+tanx)]
=e＾(-1)

追答

x->0
sinx ~ x - (1/6)x^3
cos(sinx)
~ 1 -(1/2)[x - (1/6)x^3]^2 + (1/24)[x - (1/6)x^3]^4
~ 1 - (1/2)[x^2 - (1/3)x^4] +(1/24)[ x^4]
~ 1 -(1/2)x^2 + (5/24)x^4
cosx ~1 -(1/2)x^2 +(1/24)x^4
cos(sinx) - cosx ~ (1/6)x^4
lim(x->0) [cos(sinx) -cosx ]/x^4
=lim(x->0) (1/6)x^4/x^4
=1/6
Or
lim(x->0) [cos(sinx) -cosx ]/x^4 (0/0)
=lim(x->0) [-cosx.sin(sinx) +sinx ]/(4x^3) (0/0)
分母： x^3
分子： 也要到 x^3
sin(sinx)
~ sin ( x - (1/6)x^3)
~ [ x - (1/6)x^3] - (1/6)[ x - (1/6)x^3]^3
~ x - (1/6)x^3 -(1/6)x^3
~ x - (1/3)x^3
cosx ~ 1 - (1/2)x^2
cosx. sin(sinx)
~ [ 1 - (1/2)x^2] . [x - (1/3)x^3]
~ x - (1/3)x^3 - (1/2)x^3
~ x - (5/6)x^3
sinx ~ x- (1/6)x^3
-cosx. sin(sinx) + sinx
~-[x - (5/6)x^3] +[x- (1/6)x^3]
~ (2/3)x^3
lim(x->0) [cos(sinx) -cosx ]/x^4 (0/0)
=lim(x->0) [-cosx.sin(sinx) +sinx ]/(4x^3)
=lim(x->0) (2/3)x^3/(4x^3)
=1/6

追问

第6题咋做

追答

那个，不太会

追问

好吧