prove limits...?

Question

prove Lt x->inf. cuberoot{(a+x)(b+x)(c+x)} -x is (a+b+c)/3

Answer 1

♠ let q be for cube root sign; let t=1/x → 0;
thus y(x) =q{(a+x)(b+x)(c+x)} –x;
and y(t) = [q{(a+1/t)(b+1/t)(c+1/t)} –1]/t =
and t*y(t)= q{(at+1)(bt+1)(ct+1)} –1;
♠ multiplying all under q we get:
1 +(a+b+c) t +(ab+ac+bc) t^2 +abc t^3 =
= 1 +(a+b+c)t +o(t), where o(t) << t;
♣ formula: a^3-b^3 =(a^2 +ab +b^2)*(a-b);
t*y(t) =q(1 +(a+b+c)t +o) –1 =
= [1 +(a+b+c)t +o –1] /
/ [1 +(1 +(a+b+c)t +o) +(1 +(a+b+c)t +o)^2] =
= [(a+b+c)t +o] / [2 +(a+b+c)t +o +
+1 +2((a+b+c)t +o) +((a+b+c)t +o)^2] →
→ [(a+b+c)t +o] /[3 +3(a+b+c)t +o] →
→ (a+b+c)t / 3;
♦ y(t) → (a+b+c)/3;

Answer 2

Are you sure you got that right? Because to me, it looks as if that limit diverges.