= Solution
A <continuous function> at $x_0$ satisfies the following quantified condition:
$$
\boxed{\forall\varepsilon>0\ \exists\delta>0\ \forall x\in\mathbb R:\quad |x-x_0|<\delta\ \Longrightarrow\ |f(x)-f(x_0)|<\varepsilon.}
$$
The choice of $\delta$ may depend on $x_0$ and $\varepsilon$.
For the bounded example, take
$$
\boxed{f(x)=(1-x)\sin(1/x),\qquad 0<x\leq1.}
$$
This is a <continuous function> on its domain and $|f(x)|\leq1-x<1$. Along $x_n=(\pi/2+2\pi n)^{-1}$ its values tend to $1$, while along $y_n=(3\pi/2+2\pi n)^{-1}$ its values tend to $-1$. Hence the <supremum> is $1$ and the <infimum> is $-1$, and neither is attained. “Upper and lower bound” here means the least upper and greatest lower bounds.
For the nonnegative function on the whole real line, if $f\equiv0$ the conclusion is immediate. Otherwise choose $x_*$ with $m=f(x_*)>0$. The limits at both infinities give $R>|x_*|$ such that $f(x)<m/2$ whenever $|x|>R$. By the <extreme value theorem>, a <continuous function> on the compact interval $[-R,R]$ has a maximum $M=f(\alpha)$ there. Since $M\geq m$, it also exceeds every value outside this interval. Thus \b[$f$ is bounded above and attains its global maximum]. Nonnegativity ensures that either the function is zero or such a positive comparison value exists; without it, a function such as $-e^{-x^2}$ would have an unattained <supremum> of zero.
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