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I need your help proving/disproving the following statement: if $$ f:[0,\infty) \to R $$ is continuous and $$ \lim_{x \to \infty} f(x) = L $$ then $ f $ is bounded on $ [0,\infty) $

I thought to go by definitions, but I'm not sure how exactly to connect the two.. I know there is a limit so:

For every $ \epsilon > 0 $ there exist $ M > 0 $ such that for every $ x > M $ it holds that $ \lvert f(x) - L \rvert < \epsilon $

and then the bounded defenition: There exist $ K > 0 $ such that for every $ x $ in the domain, $ \lvert f(x) \rvert < K $

So it seems true but not sure how to put it into words.. I prefer if you can help me proceed from here instead of offering your own solution.

Thank you

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    $\infty$ is not a number, and your notation should be $[0,\infty)$ (as in the answer of @Arthur) rather than $[0,\infty]$. – 2017-01-07

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Hint: You're on the right track. Here is how to continue: Divide $[0, \infty)$ into two parts: $[0, M]$ and $[M, \infty)$, and argue on each of those separately why $f$ should be bounded.

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    Thank you @Arthur , so for the $ [0,M] $ part I need to use the fact the $ f $ is continues and choose $ K = {f(x0) + 1 | x0,x (x0>x)}$ meaning choose $K$ to be the maximum of all $ f(x0) $ values? and then for the second part , $ (M,\infty) $ I can choose $ K = \epsilon + L + 1 $ and that should do the trick? – 2017-01-07
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    @Noam More or less. You have to prove that on a closed and bounded interval, the function actually has a maximum, but otherwise it's good. – 2017-01-07
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    @Noam Meaning, you can choose $K^*>0$ such that $|f(x)|\leq K^*$ for all $x\in[0,M]$. Then take $K=\max\{K^*,\epsilon+|L|\}$. – 2017-01-07
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    @Noam Just take note that if $x>M$ then $|f(x)|< \epsilon+|L|\leq K$ and if $x\in[0,M]$, then $|f(x)|\leq K^*\leq K$. Hope this helps. – 2017-01-07
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    Thank you both, understood. – 2017-01-07