Let's talk about the Lipschitz condition first (I haven't learned it, I found it online, from Baidu Encyclopedia):

Definition of Lipschitz continuity: If there is a constant k (non-negative), any two different real numbers x 1 and x2 of the domain d will have: ∣ F (X 1)-F (X2) ∣≤ K ∣ X65438.

The following proves the original proposition.

Walk in two steps.

The first step is to prove that the function f(x)/x is uniformly continuous in any closed interval [a, b].

Therefore, we first prove that the function f(x) is uniformly continuous in any closed interval [a, b].

For any given ε > 0, we say that when x 1, x2∈[a, b], ∣ x 1-x2 ∣ < ε/k, there must be.

∣ f (x1)-f (x2) ∣≤ k ∣ x1-x2 ∣1,let ε be any positive number.

Then we get that when c > max {1, 3 ∣ f (a) ∣/ε} = η 1,

∣x 1-x2∣∣f(a)∣/x 1x2

c+2- 1 = c+ 1。

This shows that both X 1 and X2 belong to [c, +∞].

In this way, we prove that the function f(x)/x is uniformly continuous on [a, +∞].

The certificate is over.