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finite limit implying uniform continuity
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(Theorem)
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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
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(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
![$\displaystyle \vert f(x_1)\!-\!f(x_2)\vert \;<\; \varepsilon \quad \forall\, x_1,\,x_2 \in [0,\,M\!+\!1]\;\;$](http://images.planetmath.org:8080/cache/objects/11874/js/img2.png "$\displaystyle \vert f(x_1)\!-\!f(x_2)\vert \;<\; \varepsilon \quad \forall\, x_1,\,x_2 \in [0,\,M\!+\!1]\;\;$") with |
(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
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(3) |
So, one of the conditions (2) and (3) is always in force, whence the assertion is true.
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"finite limit implying uniform continuity" is owned by pahio.
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Cross-references: belong, compact, function, positive, proof, uniformly continuous, number, finite, limit, interval, continuous, real function, theorem
This is version 2 of finite limit implying uniform continuity, born on 2009-08-23, modified 2009-08-23.
Object id is 11874, canonical name is FiniteLimitImplyingUniformContinuity.
Accessed 220 times total.
Classification:
| AMS MSC: | 26A15 (Real functions :: Functions of one variable :: Continuity and related questions ) |
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Pending Errata and Addenda
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