Prove that a continuous real function with finite limits is bounded $f: \mathbb{R} \rightarrow \mathbb{R}$ is a continuous function. Assume that $\lim_{x \rightarrow \pm \infty} f(x)$ exist and are finite. Prove that $f$ is bounded.
So to show that $f$ is bounded, I must show that $\exists M \in \mathbb{R}$ such that $|f(x)| \leq M$ for all $x \in Dom(f)$. Assuming that $f(x)$'s limit exist, then I know that $\forall \epsilon  > 0$, $\exists N$ such that for $n > N$, $|f(n) - L| < \epsilon$. But I am having trouble with figuring out how to connect the two ideas together. 
 A: The answer was essentially given. Here is an explicit way to connect the ideas.
Take $\epsilon=1$ in the definition of limit. So,


*

*due to $\lim_{x\to-\infty}=L_1$, there exists $N_1$ such that $f(x)\in(L_1-1,L_1+1)$ for all $x<N_1$.

*due to $\lim_{x\to\infty}=L_2$, there exists $N_2>N_1$ such that $f(x)\in(L_2-1,L_2+1)$ for all $x>N_2$.
Finally, due to continuity, $f$ is bounded on the compact interval $[N_1,N_2]$. Hence, there exists $C$ such that $|f(x)|\leq C$ for all $x\in[N_1,N_2]$.
So, you can take $M=\max\{|L_1-1|,|L_1+1|, |L_2-1|, |L_2+1|,C\}$ to conclude that $|f(x)|<M$ for all $x\in\mathbb{R}={\rm Dom}(f)$.
A: Since $\lim\limits_{\vert x\vert\to \infty}f(x)=L$, there fore for every $\epsilon>0$, there exists $M>0$ such that $\vert f(x)-L\vert< \epsilon$ for all $\vert x\vert>M$.
Also due to continuity of $f$, $f$ is bounded on $[-M,M]$
A: Hint: Take $$|f(x)|=|f(x)-L+L|\le|f(x)-L|+|L|\le\varepsilon+|L|.$$
A: You should see what goes wrong if you omit only the condition of limits at the two infinities. And then what goes wrong if you omit only the condition of continuity. That should suggest that you have to cut the problem up into 3 parts, one for the middle and one for each end. If each part is bounded, then you are done. For this to work you will need to choose where to cut based on the limiting behaviour in each direction.
