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.

Solution 1:

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 $x\in\mathbb{R}={\rm Dom}(f)$.

Solution 2:

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 $[-M,M]$