# 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 |