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Let $f(x)$ be continuous for all $\mathbb R$. $$\lim_{x\to\infty}f(x)=L_1$$ and $$\lim_{x\to-\infty}f(x)=L_2$$ Where $L_1,L_2$ belong to $\mathbb R$.

Prove that $f(x)$ is bounded for all $\mathbb R$.

My problem with this conjecture: Isn't $f(x)$=$1/x$ a counterexample? In this case, $L_1,L_2=0$ and the function is not bounded. Did I miss something?

Thanks in advance.

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Your proposed counterexample is not defined at $x=0$. – Chris Eagle Dec 29 '12 at 15:09
Your supposed counterexample isn't continuous at the origin .... – Peter Smith Dec 29 '12 at 15:10
Oops.. Thanks. How do I prove this then? – pie Dec 29 '12 at 15:10
An abstract proof, which I will not post as a solution: The existence of the two limits implies that $ f $ extends to a continuous function $ \tilde{f} $ on the two-point compactification of $ \mathbb{R} $. Hence, $ \tilde{f} $ is bounded, which implies that $ f $ is bounded as well. – Haskell Curry Dec 29 '12 at 15:46
@HaskellCurry: That is my answer. :) – tomasz Dec 29 '12 at 15:49
up vote 3 down vote accepted

Proof: $$\exists m<0: x<m\implies \left|f(x)-L_1\right|<1\iff L_1-1<f(x)<L_1+1$$ and $$\exists M>0: x>M\implies \left|f(x)-L_2\right|<1\iff L_2-1<f(x)<L_2+1$$ Now in $[m,M]$, by continuity $f$ is bounded. Can you finish this off?

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So if we take K=$max[$L_1+1,L_2 + 2$]$ to be the upper boundry and k=$min[$L_1+1,L_2 + 2$]$ to be the lower boundry, it is obvious, that for P=max[K,k], |f(x)|<P ? – pie Dec 29 '12 at 15:27
The limits of f(x) when x->-∞ and x->∞. – pie Dec 29 '12 at 15:29
My bad, meant L2+1. P.S. How do I math-type in the comments? – pie Dec 29 '12 at 15:31
Sorry, I didn't manage to activate the chat, to my emberassment. – pie Dec 29 '12 at 15:47
@pie: mathmode works in comments pretty much the same as everywhere else, just use $ signs. – tomasz Dec 29 '12 at 15:51

Hint: It is likely that you already have a theorem to the effect a function which is continuous on a closed (bounded) interval is bounded on that interval. Perhaps it is in the form that such a function attains a maximum and a minimum.

Let $A$ be a negative number such that $f(x)$ is not far from $L_2$ for $x\lt A$, and let $B$ be a positive number such that $f(x)$ is not far from $L_1$ for $x\gt B$.

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A different proof: by the existence of the limits, the function $f$ can be extended to a continuous function on $[-\infty,+\infty]$, which is compact (you can compose $f$ with $\tan$ and think instead of it as a function on $(-\pi/2,+\pi/2)$ extended to $[-\pi/2,\pi/2]$ if you're not comfortable with $[-\infty,+\infty]$).

A continuous image of a compact set is compact (or: a continuous function on a closed interval is bounded and attains its extremes, in the language of elementary analysis), so the extension of $f$ is bounded, and so is $f$.

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by existence of 2 limits,f must be bounded in some neighbourhood of infinity and minus infinity say[-infinity,a]and [b,infinity] by some M.on [a,b] again f must be bounded by continity by N say.hence f must be bounded by max(M,N)

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Given the two limits, we can find $x_1<x_2$ such that $m:=\min(L_1,L_2)-1\leq f(x) \leq \max(L_1,L_2)+1=:M$ for all $x\leq x_1$ and all $x\geq x_2$.

Now $f$ is continuous on the compact interval $[x_1,x_2]$, hence there exist $m'\leq M'$ such that $m'\leq f(x)\leq M'$ for all $x$ in $[x_1,x_2]$.

Finally, we see that $f$ is bounded above by $\max(M,M')$ and below by $\min(m,m')$ on $\mathbb{R}$.

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