if a finite limit $L = \lim_{x \rightarrow \infty} f(x) $ exists. Prove that $\exists M >0$ , $|f(x)| \leq M$ for all $x \in [a, +\infty)$

Question

The question is given below:

Let $f$ be a continuous function for $x \geq a,$ and suppose that a finite limit $L = \lim_{x \rightarrow \infty} f(x) $ exists. Prove that there exists $M >0$ such that $|f(x)| \leq M$ for all $x \in [a, +\infty).$

My Thoughts

1-But I wonder how can I prove this, will I use a proof similar to the comparison test?

2- Will the proof include the use this problem:

Let $f$ be an increasing function for $x \geq a,$ and suppose that a finite limit $L = \lim_{x \rightarrow \infty} f(x) $ exists. Prove that $f(x) \leq L$ for all $x \in [a, +\infty).$and that $f$ is bounded on $[a, +\infty).$

Any help will be appreciated.

Answer 1

By definition of limit, we know there exist a $X\in [a,+\infty)$, such that $$|f(x)-L|<1,\,\,\,x>X$$ which implies that $$|f(x)|<1+|L|,\,\,\,x>X$$ If $X=a$, let $M=\max{(1+|L|,|f(a)|)}$, then we've done.

If $X>a$, we know there exist $N>0$ such that $$|f(x)|<N,\,\,x\in[a,X]$$ by the continuity of $f$ on $[a,X]$. Let $M=\max{(1+|L|,N)}$, we've done.

Answer 2

We know that for all $\epsilon > 0$, there is an $N>0$ such that $x>N$, then $|f(x)-L| < \epsilon$.

In particular, there is an $N_1>0$, such that $x > N$, then $|f(x)-L| < 1$. then we have $|f(x) | < |L|+1$.

Let $W = \max_{x \in [a, N]} |f(x)|.$

I'm leaving the last step for you to use these to construct an upper bound for $|f(x)|$.

Remark: We can't assume that $f$ is monotonic.