Proving that a function of the form f(x+y)=f(x)*f(y) is continuous I have the to solve the following problem:
Let $f$ be a function from the real numbers to the real numbers. The function satisfies $f(x+y) = f(x)f(y)$ for all real $x,y$. Prove that if $f$ is continuous in $0$ then $f$ is continuous in every point.
I think I have a solution but I would like to know if it's correct:
By letting $y=0$ we get that $f(x)=f(x)f(0)$ meaning that $f(0)=1$. Thus we know that for every $\epsilon > 0$ and for every $x$ there exists a $\delta$ such that $|f(x)-1| < \frac{\epsilon}{|f(a)|} $ when $|x| < \delta$. Since this holds for all $x$ we can replace $x$ with $x-a$. Rewriting this we get that whenever $|x-a| < \delta$ we have that $\,|f(a)f(x-a)-f(a)|<\epsilon \implies |f(x)-f(a)| < \epsilon$.
 A: In your proof,doing $\epsilon/|f(a)|$ may be problematic as $f(a)$ could be $0$ also! You may proceed as below also. 
$f(x+y)=f(x)f(y)\implies f(0)=f^2(0)\implies f(0)=0 $ or $f(0)=1$ 
If $f(0)=0$, then $\forall x\in \mathbb R, f(x)=f(x+0)=f(x)f(0)=0$ and we are done. 
If $f(0)=1$, then for any $x_0\in \mathbb R: x_0\ne 0$:

$|f(x_0+h)-f(x_0)|=|f(x_0)(f(h)-1)|\le (|f(x_0)|+1)|f(h)-1|$. It'll be clear later why $1$ was added here. 
$\forall \epsilon\gt 0, \exists \delta\gt 0: |h|\lt \delta\implies |f(h)-1|\lt \frac{\epsilon}{(|f(x_0)|+1)}$. It follows that: 
$|f(x_0+h)-f(x_0)|\lt (|f(x_0)|+1)\times \frac{\epsilon}{(|f(x_0)|+1)}=\epsilon $ 
We have proved that: $\lim_{h\to 0} (f(x_0+h)-f(x_0))=0$ 
Now write $f(x_0+h)=(f(x_0+h)-f(x_0))+f(x_0)$. By limit rules, we have 
$\lim_{h\to 0}f(x_0+h)=0+f(x_0)=f(x_0)$ 
$x_0\ne 0$ was any arbitrary point and hence $f$ is countinuous everywhere.
A: Applying lim to both sides as y to 0 $$f(x+y) =f(x)f(y)$$
We get $$\lim_{y\to 0}f(x+y)=\lim_{y\to 0}f(x)f(y)=f(x)\lim_{y\to 0}f(y)=f(x)f(0)=f(x)$$
So by definition $f(x)$ is defined everywhere and the limit exists and is equal $f(x) $. Those 3 facts imply continuity.
A: Putting $x=y=0$, $f(0)=(f(0))^2$,   if $f(0)=0$ then for any $x \in \mathbb{R}$, $f(x)=f(x)f(0)=0$ and we are done.
So take $f(0)=1$. Then note that $1=f(0)=f(x)f(-x)$ holds for any $x \in \mathbb{R}$ which implies $f(x)\neq 0$ for all $x \in \mathbb{R}$ in this case. Hence your proof is perfectly right.
Sequence can also be used to prove this: $\lim_{n \rightarrow \infty} x_n=a\Rightarrow \lim_{n \rightarrow \infty} (x_n-a)=0\Rightarrow \lim_{n \rightarrow \infty} f(x_n-a)=f(0)=1 \Rightarrow \lim_{n \rightarrow \infty}f(x_n)f(-a)=1 \Rightarrow \lim_{n \rightarrow \infty} f(x_n)=\frac{1}{f(-a)}=f(a)$.
