Question

谁能帮我解出1-4题的过程

Answer 1

(1)
f(x)= (x+1)/(x^2+1)
f(x/2)
= (x/2+1)/[(x/2)^2+1]
= 2(x+2)/(x^2+4)
(2) lim(x->∞) sinx/x =0
(3) lim(x->0) sinkx/x = lim(x->0) kx/x =k
(4)
lim(x->∞) (2x^2+2)/(3x^3 -1)
= lim(x->∞) (2/x+2/x^3)/(3 -1/x^3)
=0
(5)
lim(x->∞) (2x^2+2)/(3x^2 -1)
=lim(x->∞) (2+2/x^2)/(3-1/x^2)
=2/3
(6)
(x+2)(x-1)=0
x=1 or -2
间断点: x=1 , x=-2
(8)
f(x)
=x^2 ; x≤0
=x+a ; x>0
f(0-)=f(0) =0^2=0
f(0+)=lim(x->0) (x+a) = a
=> a=0
(9)
y=x^3
y'=3x^2
y'|x=1 = 3
切线方程
y-1= 3(x-1)
3x-y -2 =0

Answer 2

如图