Question

谁能帮我做几道函数题啊 ？ 急.急.急a

Answer 1

1)f(sinx/2)=cosx+1=1-2sin^2(x/2)+1=2-2sin^2(x/2)
f(cosx/2)=2-2cos^2(x/2)=1-(2cos^2(x/2)-1)=1-cosx
2)t=lnx x=e^t
f(x)=e^(2t)(1+x)
3-1)=lim[sinx^n/x^n]/(tan^nx/x^n)
=lim[sinx^n/x^n]/(tanx/x)^n=1/1^n=1
3-2)=lim(1+5/(x-3))^x=lim(1+5/(x-3))^(x-3+3)/5*5
=lim[(1+5/(x-3))^(x-3)/5]^5*[(1+5/(x-3))]^3
=e^5*1=e^5
3-3)
3-3)=lim(tanx-sinx)/[sin^3x(√(1+tanx)+√(1+sinx))
=1/2lim(sinx/cosx-sinx)/sin^3x
=1/2lim(secx-1)/sin^2x
=1/2lim(secxtanx/2sinxcosx)
=1/4limsec^3x=1/4

追问

亲，你知道第4题吗

追答

4-1) x^2+x-6=(x+3)(x-2)=0
x=-3 x=2 判断在这2点是否连续
f(x)=(x+3)(x+1)(x-1)/[(x+3)(x-2)] x=-3连续 x=2断
4-2)lim h(x)=1/2
x→0+
lim h(x)=x/{x[√2+√(2-x)]}=√2/4 x=0不连续
x→0-

追问

谢谢啊