# finite limit implying uniform continuity

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

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

 $\displaystyle|f(x)\!-\!a|\;<\;\frac{\varepsilon}{2}\quad\forall x>M.$ (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  $\delta<1$  such that

 $\displaystyle|f(x_{1})\!-\!f(x_{2})|\;<\;\varepsilon\quad\forall\,x_{1},\,x_{2% }\in[0,\,M\!+\!1]\;\;\mbox{with}\;\;|x_{1}\!-\!x_{2}|<\delta.$ (2)

Let $x_{1},\,x_{2}$ be nonnegative numbers and  $|x_{1}\!-\!x_{2}|<\delta$.  Then  $|x_{1}\!-\!x_{2}|<1$  and thus both numbers either belong to  $[0,\,M\!+\!1]$  or are greater than $M$.  In the latter case, by (1) we have

 $\displaystyle|f(x_{1})\!-\!f(x_{2})|\;=\;|f(x_{1})\!-\!a\!+\!a\!-\!f(x_{2})|\;% \leqq\;|f(x_{1})\!-\!a|+|f(x_{2})\!-\!a|\;<\;\frac{\varepsilon}{2}+\frac{% \varepsilon}{2}\;=\;\varepsilon.$ (3)

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

Title finite limit implying uniform continuity FiniteLimitImplyingUniformContinuity 2013-03-22 19:00:20 2013-03-22 19:00:20 pahio (2872) pahio (2872) 5 pahio (2872) Theorem msc 26A15