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Let $a\in\Bbb{R}$. Suppose $f$ is a real valued continuous function on $[a, \infty)$ satisfying that $\lim\limits_{x\to\infty}f(x)=L\in\Bbb{R}$ I need to show:

$f$ is bounded on $[a, \infty)$

Here is what I have so far: if $f$ is bounded on $[a, \infty)$, then there exists a constant $M$ s.t. $|f(x)| \le M$ for all $x \in [a, \infty)$

From the fact that $\lim\limits_{x\to\infty} f(x) = L$, I can say that given $\epsilon > 0$, there is a $\delta$ st $|f(x) - L| < \epsilon$.

If $L=0$, then $|f(x)| < \epsilon$. I can set $M \le \epsilon$, but I don't think this last part makes any sense because there is no way I can claim that $L =0$

Or what if I say that since $|f(x) - L| < \epsilon$, then $|f(x) - L + L| < \epsilon + |L|$. If I let $M = \epsilon + |L|$, then would it work? Please provide me with some hints

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    [Related](http://math.stackexchange.com/q/527250/28900).2013-10-15

1 Answers 1

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The continuity condition is quite essential, here. Consider the function $f(x)=\begin{cases}\frac1x & x\neq 0\\0 & x=0.\end{cases}$

Then $\lim\limits_{x\to\infty}f(x)=0$, but $f$ is certainly not bounded on $[a,\infty)$ for $a=0$.


Set $\epsilon>0$, so that you know there is some $N>a$ such that $|f(x)-L|<\epsilon$ for $x>N$. (Note the difference between what I've put here and your statement with the $\delta$.) As you noted, we can say that $|f(x)|\le|L|+\epsilon$ for certain $x$--in particular, for $x\in(N,\infty)$.

Now, $f$ is continuous on the compact interval $[a,N]$, so is bounded--meaning that there is some $M_0$ such that $|f(x)|\le M_0$ for all $x\in[a,N]$. Thus, putting $M=\max\{|L|+\epsilon,M_0\}$, it follows that $|f(x)|\le M$ for all $x\in[a,\infty)=[a,N]\cup(N,\infty)$.

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    That was a typo. I'll fix it.2012-12-06