Suppose that $f: [0,\infty) \to \mathbb{R}$ is a continuous function and $\displaystyle \lim_{x \to \infty}f(x)$ exists. Prove that $f$ is uniformly continuous on $[0,\infty)$.
Here's my solution, but because the problem has been given in a math contest preparation program I'm afraid that there's some point that I might be missing.
Since $\displaystyle \lim_{x \to \infty}f(x)$ exists, there exists $L \in \mathbb{R}$ such that:
$\forall \epsilon>0, \exists M>0 \text{ such that } x > M \implies |f(x)-L|<\epsilon/2$
$\forall \epsilon>0, \exists M>0 \text{ such that } y > M \implies |f(y)-L|<\epsilon/2$
Therefore, $\forall x,y \in (M,\infty)$ we have shown that:
$|f(x)-f(y)|<\epsilon$.
Now, let's consider $f$ on $[0,M]$. Since $[0,M]$ is a compact interval and $f$ is continuous, $f$ is uniformly continuous on $[0,M]$:
$\forall \epsilon>0, \exists \delta_1>0, \forall x,y \in [0,M]: |x-y|<\delta_1 \implies |f(x)-f(y)|<\epsilon$.
Now let us focus our attention at $x=M$.
Since $f: [0,\infty) \to \mathbb{R}$ is continuous, it is continuous at $x=M$ as well. Therefore:
$\epsilon>0, \delta_2>0, \forall x: |x-M|<\delta_2 \implies |f(x)-f(M)|<\epsilon/2$
So, for any $x,y \in (M-2\delta_2,M+2\delta_2)$ we have $|f(x)-f(y)|<\epsilon$
Now if we set $\delta=\min\{\delta_1,\delta_2\}$ we see that all the $3$ cases above become true and we conclude that $\forall x,y \in [0,\infty): |x-y|<\delta \implies |f(x)-f(y)|<\epsilon$. This proves that $f$ is uniformly continuous on $[0,\infty)$.
Is my nonsense considered a proof?
Answer
Did you make a mistake in ".. for any $x, y \in (x-2\delta_2, x+2\delta_2)$..."? Because M'd disappeared. But I think that you are almost there.
It worth trying this with sequence. If the result is false, then there is $\epsilon>0$ and sequences $x_n$ and $y_n$ such that
$$|x_n - y_n| < 1/n$$
and
$$|f(x_n)-f(y_n)|\geq \epsilon.$$
The hypothesis shows that $y_n$ and $x_n$ are bounded, and Bolzano Weierstrass gives a contradiction.
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