Suppose $f$ is continuous on $[a, \infty)$ and that the limit ( as $x$ approaches infinity) $=L$ for some real number $L$. Prove that $f$ is uniformly continuous on $[a, \infty)$. Please ask if you have any questions about this problem.

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Downvoted because you already asked a similar question and have not provided any background work on that question. We're not here to do your homework. –  Arkamis Jul 19 '12 at 19:59
What have you tried? –  Code-Guru Jul 19 '12 at 20:06
fn(x) as n tends to infinity = f |fn(x) - f | < E |f(x) - L | < E |L + h - L | < E h < E –  Rocket 12 Jul 19 '12 at 20:10
@AdriánBarquero He probably means "if anything is unclear let me know". –  Pedro Tamaroff Jul 19 '12 at 20:37

Here's an outline of one approach:

$\ \ \ 1)$ Let $\epsilon>0$.

$\ \ \ 2)$ Choose $N$ so that $|f(x)-L|<\epsilon/3$ whenever $x\ge N$.

$\ \ \ 3)$ Argue that $f$ is uniformly continuous on $[a,N]$.

$\ \ \ 4)$ Choose $\delta>0$ so that $|f(x)-f(y)|<\epsilon/3$ whenever $|x-y|<\delta$ and $x\in[a,N]$, $y\in[a,N]$.

$\ \ \ 5)$ Show that in fact $|f(x)-f(y)|<\epsilon$ whenever $|x-y|<\delta$ and $x\in[a,\infty)$, $y\in[a,\infty)$

$\ \ \ \ \ \ \$(consider three cases depending on the relationship between $x$, $y$, and $N$).

A slightly more elegant approach:

$\ \ \ 1)$ Let $\epsilon>0$.

$\ \ \ 2)$ Choose $N$ so that $|f(x)-L|<\epsilon/2$ whenever $x\ge N$.

$\ \ \ 3)$ Argue that $f$ is uniformly continuous on $[a,N+1]$.

$\ \ \ 4)$ Choose $1>\delta>0$ so that $|f(x)-f(y)|<\epsilon$ whenever $|x-y|<\delta$, $x\in[a,N+1]$, and

$\ \ \ \ \ \$$y\in[a,N+1]$.

$\ \ \ 5)$ Show that in fact $|f(x)-f(y)|<\epsilon$ whenever $|x-y|<\delta$ and $x\in[a,\infty)$, $y\in[a,\infty)$

$\ \ \ \ \ \ \$(consider two cases depending on whether one (or both) of $x$, $y$, exceeds $N+1$).

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