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Let $f$ be a $mathbb R rightarrow mathbb R$ continuous function such that : $lim_ x to pm infty f(x) in mathbb R$ and $lim_ x to 0 f(x) in mathbb R$

How can one show that $f$ is bounded ? I get it "intuitively" but I cant show it rigorously

The fact that $lim_xto0f(x)inmathbb R$ is necessarily true: since $f$ is continuous that limit has to be $f(0)$$.

Now, suppose that $f$ is unbounded. Then, for each $ninmathbb N$, there is a $x_ninmathbb R$ such that $bigllvert f(x_n)bigrrvertgeqslant n$. The sequence $(x_n)_ninmathbb N$ is either bounded or unbounded and:

• if it is bounded, then it has a convergent subsequence $(x_n_k)_kinmathbb N$. But if $lim_ktoinftyx_n_k=x$, then $lim_ktoinftyf(x_n_k)=f(x)$, which is impossible, since $bigl(f(x_n_k)bigr)_kinmathbb N$ is unbounded.


• if it is unbounded, then it has a subsequence $(x_n_k)_kinmathbb N$ whose limit is $pminfty$, and we then get a similar contradiction.

If $lim_xto-infty f(x)=a$ and $lim_xtoinfty f(x)=b$ put $|a|+|b|+1=:c$. There is an $M>0$ such that $|f(x)|leq c
$ for all $xgeq M$ and all $xleq-M$. Since $f$ is continuous there is a $c'$ such that $|f(x)|leq c'$ for all $xin[-M,M]$. It follows that $|f(x)|leq c+c'$for all $xinmathbb R$.

Juts for amusement:

Let $phi(x) = begincases lim_x to -infty f(x) , & x = -pi over 2 \
f(tan(x)), & x in (-pi over 2, pi over 2 ) \
lim_x to +infty f(x) , & x = pi over 2
endcases $.

Show $phi$ is continuous on the compact set $[-pi over 2,pi over 2]$, hence bounded, and hence $f$ is bounded.

1) Consider $[0,infty)$.

$lim_x rightarrow inftyf(x)=L:$

For $epsilon >0$ there is a $M$, real, positive, s.t.

for $x > M$ $|f(x)-L| <epsilon$, i.e.

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

The continuous function $f$ attains minimum and maximum on the compact interval $[0,M]$.

Hence $f$ is bounded on $[0,infty)$.

2) Proceed likewise for $(-infty,0]$.

1) and 2): f is bounded on $mathbbR$.

Also cf.. comment of miraunpajaro

The hypothesis that

$displaystyle lim_x to infty f(x) in Bbb R tag 1$

means that

$exists L_+ in Bbb R, ; displaystyle lim_x to infty f(x) = L_+; tag 2$

that is,

$forall 0 < epsilon_+ in Bbb R ; exists 0 < M_+ in Bbb R, ; x > M_+ Longrightarrow vert f(x) - L_+ vert < epsilon_+; tag 3$

that is,

$x > M_+ Longrightarrow L_+ - epsilon_+ < f(x) < L_+ + epsilon_+; tag 4$

likewise the hypothesis

$displaystyle lim_x to -infty f(x) in Bbb R tag 5$

gives us

$exists L_- in Bbb R, ; displaystyle lim_x to -infty f(x) = L_-; tag 6$

i.e.,

$forall 0 < epsilon_- in Bbb R ; exists 0 > M_- in Bbb R, ; x < M_- Longrightarrow vert f(x) - L_- vert < epsilon_-, tag 7$

or

$x < M_- Longrightarrow L_- - epsilon_- < f(x) < L_- + epsilon_-; tag 8$

it follows that, letting

$b = min(L_- - epsilon_-, L_+ -epsilon_+), ; B = max(L_- + epsilon_-, L_+ + epsilon_+) tag 9$

and

$m = max(-M_-, M_+), tag10$

that

$vert x vert > m Longrightarrow b < f(x) < B, tag11$

so $f(x)$ is bounded on the set $(-infty, m) cup (m, infty)$; furthermore, since the closed interval $[-m, m]$ is compact and $f(x)$ is continuous, $vert f(x) vert$ is strictly bounded on this interval by some $0 < beta in Bbb R$:

$x in [-m, m] Longrightarrow -beta < f(x) < beta; tag12$

combining (11) and (12) shows that $f(x)$ is bounded on all of $Bbb R$. Indeed, we have

$forall x in Bbb R, ; vert f(x) vert < max(vert b vert, vert B vert, beta). tag13$