Proof that a function is bounded

Question

The question :

Let $f:[1,\infty)\to \mathbb{R}$ be a continuous function such that $\underset{x\rightarrow\infty}{\lim}f(x)=L$

Prove that the function is bounded.

My try :

By definition a continuous function $f:[1,\infty)\rightarrow\mathbb{R}$ 1. $f$ is continuous in $(1,\infty)$ i.e $\forall x_{0}>1\underset{x\rightarrow x_{0}}{\lim}f(x)=f(x_{0})$

2. $f$ continuous at $1^{+}$ i.e $\underset{x\rightarrow1^{+}}{\lim}f(x)=f(1)$

as stated in the question's contitions :$ \underset{x\rightarrow\infty}{\lim}f(x)=L\iff\forall\varepsilon>0\,\exists M\in\mathbb{R}: x>M\rightarrow\left|f(x)-L\right|<\varepsilon$

Set $\varepsilon=\left|f(1)-L\right|$

Then exists $M$ such that $x>M\rightarrow\left|f(x)-L\right|\leq\left|f(1)-L\right|$ including when $x=1$

Set $\left|f(1)-L\right|=K$

This also implies that$ M\geq1$

Thus we get $x\geq1\rightarrow\left|f(x)-L\right|\leq K \iff x\geq1\rightarrow L-K\leq f(x)\leq L+K$

Thus $f$ is bounded.

However I feel that this proof doesnt work and I do not fully understand what I have done here actually (just tried to replicate the lecture notes of my professor)

I want to understand this question and the correct way to answer it. Assistance will be greatly appreciated.

Answer 1

Since the comments seem to indicate that the OP is not familiar with the theorem that a continuous function $f$ is bounded on a compact set, here is a simple proof for the special case of a closed bounded interval $[a,b] \subset \mathbb R$.

Suppose for a contradiction that $f$ is not bounded on $[a,b]$. Therefore it must be either unbounded above or unbounded below. Without loss of generality, assume $f$ is unbounded above. (Otherwise replace $f$ with $-f$.)

Now divide the interval into two subintervals, $[a + (a+b)/2]$ and $[(a+b)/2, b]$. Now $f$ must be unbounded on one of these subintervals. Repeating this procedure, we identify a sequence of closed bounded intervals $[a,b] = I_0 \supset I_1 \supset I_2 \supset \cdots$ such that the length of $I_n$ is $(b-a)/2^n$, and $f$ is unbounded on each of these intervals.

This means that we can choose points $x_0 \in I_0$, $x_1 \in I_1$, $x_2 \in I_2$, etc. such that $f(x_n) > n$ for every $n$.

Now each $I_n$ contains every $x_k$ for $k \geq n$, and from this we can easily conclude that $x_k$ converges to some limit $x$. Since each $x_n$ is in $[a,b]$ and $[a,b]$ is closed, it contains all of its limit points, hence $x\in [a,b]$.

By continuity of $f$, we must have $$\lim_{n \to \infty}f(x_n) = f(x)$$ but this is impossible since $f(x_n) > n$ for every $n$.

Our assumption that $f$ is unbounded on $[a,b]$ is untenable, so $f$ must be bounded after all.

Answer 2

Take $\epsilon=1$ then there is $N$ so that if $x>N$ we have $|f(x)-L|<1$.

So the function is bounded on $[N,\infty)$.

On the other hand, every continuous function defined on a closed bounded interval is bounded. So $f$ is bounded on $[1,N]$.

Since $f$ is bounded on $[1,N]$ and $[N,\infty)$ we have $f$ is bounded on $[1,\infty)$ as desired.