## You are here

Homefinite limit implying uniform continuity

## Primary tabs

# 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 force, whence the assertion is true.

## Mathematics Subject Classification

26A15*no label found*

- Forums
- Planetary Bugs
- HS/Secondary
- University/Tertiary
- Graduate/Advanced
- Industry/Practice
- Research Topics
- LaTeX help
- Math Comptetitions
- Math History
- Math Humor
- PlanetMath Comments
- PlanetMath System Updates and News
- PlanetMath help
- PlanetMath.ORG
- Strategic Communications Development
- The Math Pub
- Testing messages (ignore)

- Other useful stuff
- Corrections