Now do a deep breath and dive into the definition of limits. From last equality we have. Vε > 0 : 9X â R : Vx>X : |f(x) - C| < ε. From equality lim xââ g(x) > -1,.
Let’s consider first equality xf 0 (x) =0 x→∞ f (x) lim
and rewrite it as follows lim x[ln(f (x))]0 = 0.
x→∞
It’s ok to write natural logarithm, since f : R → R+ . Last equality leads us to following conclusion lim [ln(f (x))]0 = 0,
x→∞
otherwise previous equality cannot be correct. In its turn this means lim ln(f (x)) = K < +∞,
x→∞
and due to natural logarithm properties leads to lim f (x) = C < +∞.
x→∞
Now do a deep breath and dive into the definition of limits. From last equality we have. ∀ε > 0 : ∃X ∈ R : ∀x > X : |f (x) − C| < ε. From equality lim g(x) > −1,
x→∞
we have ∃k > −1 : ∃Y ∈ R : ∀x > Y : g(x) ≥ k > −1. Now should be obvious, that if
X x > Z = max X, Y, 1+k
than x + xg(x) > X. Now we know that ∀ε > 0 : ∃Z ∈ R : ∀x > Z : |f (x + xg(x)) − C| < ε, and remember ∀ε > 0 : ∃X ∈ R : ∀x > X : |f (x) − C| < ε, which means ∀ε > 0 : ∃X ∈ R : ∀x > X : |f (x) − f (x + xg(x))| < 2ε. This means both functions f (x) and f (x + xg(x)) are “almost equal” for “big enough” values of x. They cannot be zero (remember, f : R → R+ ) so we immediately conclude f (x + xg(x)) = 1. lim x→∞ f (x)
