Proof that if the limit exists, then the function is bounded in some neighbourhood

Question

So, here's the theorem that I'm trying to prove:

Let $f$ be a function. Then, if $f(x) \to L$ as $x \to x_0$, where $L,x_0 \in \mathbb{R}$, then $f(x)$ is bounded in some deleted neighbourhood of $x_0$.

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Proof Attempt:

Since $f(x)$ has a limit at $x = x_0$, we have:

$$\forall \epsilon > 0 : \exists \delta > 0: 0 < |x-x_0| < \delta \implies |f(x) - L| < \epsilon$$

Let $\epsilon > 0$. Then, we are guaranteed the existence of a deleted neighbourhood of $x_0$. So, we have:

$$|f(x)-L| \geq | |f(x)|-|L| |$$

$$|f(x)-L| \geq |f(x)|-|L|$$

$$|f(x)|-|L| < \epsilon$$

$$|f(x)| < |L| + \epsilon$$

Now, let $M = |L| + \epsilon$, so that $f(x) \leq M$. We have found a real number that satisfies the condition for a function to be bounded in this deleted neighbourhood. Hence, we conclude that $f(x)$ is bounded in some deleted neighbourhood of $x_0$.

In fact, I believe that a stronger result can be given here. Based on the argument above, it seems like this should hold for any deleted neighbourhood of $x_0$. Could someone have a look at this argument and tell me if it is correct and if the generalization I'm proposing does, indeed, hold? Thanks!

Answer 1

Your proof is correct.

Your generalization is not. In the proof you claim correctly that

we are guaranteed the existence of a deleted neighbourhood

in which the inequalities you state involving $\epsilon$ are true.

That argument applies only to this particular neighborhood, not to all neighborhoods.

If your argument were correct you could use the whole space as a neighborhood and conclude that the function was a bounded function.

Answer 2

This is quasi immediate. Choose any $\epsilon$, then it is guaranteed that there is a $\delta$-neighborhood where

$$L-\epsilon<f(x)<L+\epsilon.$$

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As regards the generalization, a counterexample is given by $f(x)=x$.

Taking $x_0=0$,

$$|x|<\delta\iff |x|<\epsilon$$

and $\delta=\infty$ (i.e. unbounded $x$) cannot be achieved by a finte $\epsilon$.