I need to prove the following:

Assume that $f: \mathbb{R} \to \mathbb{R}$ is such that $f(x+y)=f(x)f(y)$ for all $x,y \in \mathbb{R}$. If $f$ has a limit at zero, prove that $f$ has a limit at every point and either $\lim\limits_{x\rightarrow 0}f(x)=1$ or $f(x)=0$ for all $x \in \mathbb{R}$.

My attempt:

Since $f$ has limit at zero we have that given $\epsilon>0$ exists a $\delta >0$ such that:

$$|x|<\delta \Rightarrow |f(x)-L|<\epsilon$$ Now consider an arbitrary point $x_0 \mathbb{R}$, since we know that every real number is an acumulation point of the real numbers we have that $x_0$ is an accumulation point of the dominion of $f$, and we have the next results:

1)Let $f : D\to \mathbb{R}$ be a function and $x_0 \in \mathbb{R}$ be an accumulation point of $D$. $f$ has limit in $x_0$ iff for all sequence $\{x_n \}$ with $x_n \in D-\{x_0\}$ that converge to $x_0$, the sequence $\{f(x_n) \}$ is convergent.

2)Let $A \subset \mathbb{R}$, $a_0 \in \mathbb{R}$ an accumulation point of $A$ iff exists a sequence $\{x_n \} \in A$ with $x_n$ diferent from $a_0$ for all n such that $\{x_n\}$ converges to $a_0$

then we have that $f$ has limit in every point. My question is Am I right? an How Can I prove the other part ? Thank you.

  • $\begingroup$ Does it say that the function is continuous? $\endgroup$
    – user119615
    Sep 14, 2014 at 21:56
  • $\begingroup$ No :) It is evereything :) $\endgroup$
    – user162343
    Sep 14, 2014 at 21:57

1 Answer 1


Here is my attemt:

Just first about what it actually means to have a limit at every point. I am not sure what this means, but I assume it means that if f has a limit at every point then for every point x: $\lim_{x \rightarrow a}f(x)=L_a$. This in turn means that for any $\epsilon_a$ and a, there is a $\delta_a$ so that if $|x-a|<\delta_a$, then $|f(x)-L_a|<\epsilon_a$, this in turn means that f is continuous and that $f(a)=L_A$

1) First to prove that f has a limit at every point, I interpret this to prove that $lim_{x \rightarrow a}f(x)$ exists for every a, given that it exists for 0.

We have:

$\lim_{x \rightarrow a}f(x)=\lim_{h \rightarrow 0}f(a+h)=\lim_{h \rightarrow 0}f(a)f(h)=f(a)*\lim_{h \rightarrow 0}f(h)$. Since we know that the expression on the right exists, you can go backwards to see that the expression on the left also exists.

2) To prove the second statement assume that $\lim_{x \rightarrow 0}f(x) \ne 1$, then we must show that f(x) is zero for all x.

$f(x)=f(x-h)*f(h)$, then let h go to zero, and use my explanation above that f is continuous, and part 1 that the limit exists in every point. Then we get $f(x)=\lim_{h \rightarrow 0}f(x-h)*\lim_{h \rightarrow 0}f(h)=f(x)*\lim_{h \rightarrow 0}f(h)$. And since we know that the last limit is not 1, f(x) must be zero.

  • $\begingroup$ thank you, and yes the limit has to exists for every a like you said :) but am I done with this or I have to do more math to get to the proof? $\endgroup$
    – user162343
    Sep 14, 2014 at 22:32
  • $\begingroup$ @user162343 What do you mean more math? Are you asking if this proves the second statement? $\endgroup$
    – user119615
    Sep 14, 2014 at 22:35
  • $\begingroup$ Can you explain me what do you mean to go backwards? in the first part because I did not understand the proof :) thank you $\endgroup$
    – user162343
    Sep 14, 2014 at 22:36
  • 1
    $\begingroup$ The second statement is to prove that either the limit at 0 is 1 OR, that f(x) is zero for all x. When you are going to prove A or B, you can prove it by assuming that A is false and show that then B is forced to be true, this is taught in some rules of logic. $\endgroup$
    – user119615
    Sep 14, 2014 at 22:42
  • 1
    $\begingroup$ @user162343 You're welcome, but please check that the definition of what it means that "f has a limit at every point", or "f has a limit at 0". I assumed it meant what I wrote at the start of my post, but I think it may also mean that the limit exists for points arbitrary close to 0 for instance, but not at the point 0 itself. And if that is what is meant, then my proof is not correct, because then we are not able to get the continuity at the start. So please check what is meant in your exercise. $\endgroup$
    – user119615
    Sep 14, 2014 at 23:03

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