Let $f : [0, \infty) \to R$ be continuous such that $lim_{x \to +\infty} f(x) = 0$.
How can I Prove that f is bounded on $[0, \infty)$.
I know that condition for a function to be bounded is - There exists a real number M such that $|f(x)| \leq M$ , for all x in X(domain).
If ...the intention was to take the limit when $\;x\to 1^-\;$ , then:
Take, say $\;\epsilon=0.1\;$ , and since $\;f(x)\xrightarrow[x\to 1^-]{}0\;$ , there exists $\;\delta>0\;$ s.t
$$1-x<\delta\implies |f(x)|<\epsilon=0.1$$
But also $\;f\;$ is bounded on $\;[0\,,\,\,1-\epsilon=0.9]\;$ by Weierstrass thorem, thus all in all...
For $\;x\to \infty\;$ As above, there exists $\;R\in\Bbb R^+\;$ s.t.
$$x>R\implies |f(x)|<\epsilon$$
and $\;f\;$ is bounded in $\;[0,R]\;$ ...