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

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.

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 $$|f(x)| by the continuity of $$f$$ on $$[a,X]$$. Let $$M=\max{(1+|L|,N)}$$, we've done.

• Could you please clarify from where your forth line come ? – hopefully Dec 7 '18 at 7:55
• $|f(x)-L|<1$ is also $-1<f(x)-L<1$ and $-1+L<f(x)<1+L$. Then $$|f(x)|<\max(|L-1|,|L+1|) \leq 1+|L|$$ – Lau Dec 7 '18 at 7:58
• is this a well known fact? – hopefully Dec 7 '18 at 8:22
• I think so ..... – Lau Dec 7 '18 at 8:24
• @hopefully edited – Lau Dec 7 '18 at 9:05

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.

• I think $L+1$ is not necessary a positive number. – Lau Dec 7 '18 at 6:08
• @hopefully the answer has actually been provided in the other answer but the idea is that we know number that are big enough is bounded by a number $M_1$, number that are not too large is bounded by $M_2$. how do you select a common upper bound for both smaller number and big number at the same time? A simple analogy, you have two children, one spends more than the other, say one spend $5$ dollars per day and one spend $10$ dollars per day? how do you make sure they have enough to spend each day if you want to give them the same amount to spend. – Siong Thye Goh Dec 7 '18 at 6:28