Let f be a continuous R-valued function defined on R. Let $f$ be a continuous $\mathbb{R}$-valued function defined on $\mathbb{R}$. Assume that the limits $$L' := \lim_{x\to-\infty} f(x)$$
and $$L := \lim_{x\to+\infty} f(x)$$ both exist. 
Consider the following cases: $L' < L, L' > L$ and $L = L'$
.
Prove or disprove the following statements for the above cases:
(i) $f$ attains at least one of its maximal values or minimum value.
(ii) $f$ attains its maximal values and its minimum value.
(iii $f$ is uniformly continuous on R.
 A: A rule of thumb is that for functions that aren't continuous, there should be very little that is guaranteed about the function.
Since $f$ doesn't need to be continuous, you can pick the values of $f$ for particular $x$s however you like. You should be able to make (i), (ii), (iii) all false. For example, to make (i) false, try to make $f$ attain larger and larger values. E.g. you could make $f(1) = 1$, $f(1/2) = 2$, $f(1/3) = 3$, and so on, so that $f$ attains arbitrarily large values.
A: Here is a sketch that may help you.
We have: $$\begin{align*}L' &:= \lim_{x\to-\infty} f(x) \\ L &:= \lim_{x\to+\infty} f(x) \end{align*}$$
and that $f$ is continuous. 
Hence we find an $M \in \Bbb{R}$ s.t. $$\begin{align*}f(x) &\in [L'-\varepsilon,L'+\varepsilon] &\quad& \text{ for } x < -M \\ f(x) &\in [f_\min, f_\max] &\quad& \text{ for } x \in [-M,M] \\ f(x) &\in [L-\varepsilon,L+\varepsilon] &\quad& \text{ for } x > -M\end{align*} \\$$
where $f_\min, f_\max$ are the minimum / maximum value from $f$ on $[-M,M]$.The first and the last condition are because of the existing limits, the middle one because of $f$ is continuous.
This and $$f(x) = \arctan(x)$$ should help you to answer i) - iii)
