Prove that a function has limit everywhere.

Question

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.

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.