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

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

  • $\begingroup$ Could you provide some details about this last step please? $\endgroup$ – Secretly Dec 7 '18 at 6:03
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    $\begingroup$ I think $L+1$ is not necessary a positive number. $\endgroup$ – Lau Dec 7 '18 at 6:08
  • $\begingroup$ true, thanks for the feedback. $\endgroup$ – Siong Thye Goh Dec 7 '18 at 6:24
  • $\begingroup$ @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. $\endgroup$ – Siong Thye Goh Dec 7 '18 at 6:28
  • $\begingroup$ Could you please clarify from where your third line come? $\endgroup$ – Secretly Dec 7 '18 at 7:57

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