# Prove this conjecture - is it even possible?

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?

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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

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?
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
@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$. - A different proof: by the existence of the limits, the function$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$. - by existence of 2 limits,f must be bounded in some neighbourhood of infinity and minus infinity say[-infinity,a]and [b,infinity] by some M.on [a,b] again f must be bounded by continity by N say.hence f must be bounded by max(M,N) - Given the two limits, we can find$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}\$.