finite limit implying uniform continuity

Theorem.  If the real function $f$ is continuous on the interval  $[0,\mathrm{\infty })$  and the limit  $\underset{x\to \mathrm{\infty }}{lim}f(x)$  exists as a finite number $a$, then $f$ is uniformly continuous on that interval.

Proof.  Let  $\epsilon >0$.  According to the limit condition, there is a positive number $M$ such that

(1)

The function is continuous on the finite interval  $[0,M+1]$;  hence $f$ is also uniformly continuous on this compact interval.  Consequently, there is a positive number    such that

(2)

Let $x_1,x_2$ be nonnegative numbers and  .  Then    and thus both numbers either belong to  $[0,M+1]$  or are greater than $M$.  In the latter case, by (1) we have

(3)

So, one of the conditions (2) and (3) is always in , whence the assertion is true.

Title	finite limit implying uniform continuity
Canonical name	FiniteLimitImplyingUniformContinuity
Date of creation	2013-03-22 19:00:20
Last modified on	2013-03-22 19:00:20
Owner	pahio (2872)
Last modified by	pahio (2872)
Numerical id	5
Author	pahio (2872)
Entry type	Theorem
Classification	msc 26A15