Given a set of conditions for $f$, prove $f$ is continuous $\forall x \in \mathbb{R}$ 
Problem: 
  Let $f(x)$ be a function whose domain is $\mathbb{R}$. It is known that
  
  
*
  
*$f(x)$ is continuous at $0$
  
*$f(x+y) = f(x)f(y)$ $\ \ \forall \ x, y \in \mathbb{R}$
  
  
  Show that $f(x)$ is continuous in $\mathbb{R}$

My Attempted Solution:
$$
\begin{equation}
\begin{split}
f(x+y) &= f(x)f(y) \\ \\
\implies f(x) &= \frac{f(x+y)}{f(y)} \\ \\
\therefore \lim_{x \to 0} \ f(x) &= f(0) \\ \\
\implies \frac{f(0+y)}{f(y)} &= f(0) \\ \\
\implies f(0) &= 1
\end{split}
\end{equation}
$$
All I've done is shown $f(0) = 1$. But I am unsure how to proceed. Any suggestions or possible solutions? 
Also correct me if I'm wrong, but it would be completely circular to use the limit laws in this situation, as the direct substitution property which will be needed to evaluate the limits, require the function you're taking the limit of to be continuous in the first place.
 A: First, $f(0 + 0) = (f(0))^2$ gives $f(0) = 0$ or $1$.
If $f(0) = 0$, then $f(x) = 0$, for all $x \in \Bbb R$ and there's nothing to prove.
If $f(0) = 1$, then for all $x \in \Bbb R$,
$$\lim_{h \to 0} f(x+h) = \lim_{h \to 0} f(x)f(h) = f(x) \lim_{h\to 0}f(h) = f(x)$$
Which shows that $f$ is continuous 
A: You can show that $f(0)=1$ by substituting $y=0$ in the second equivalence of your hypothesis. Indeed, let $f(0)=a$. Then, we have
$$f(x+y)=f(x)f(y)\Rightarrow f(x+0)=f(x)f(0)\Rightarrow \\f(x)=af(x)\Rightarrow f(x)(a-1)=0\Rightarrow a=1$$
Now, we wish to show that $\lim_{x\rightarrow x_0}f(x)=f(x_0)$. 
Equivalently, for $h=x-x_0$
$$\lim_{h\rightarrow o}f(h+x_0)=\lim_{h\rightarrow 0}f(h)f(x_0)=f(x_0)\lim_{h\rightarrow 0}f(h)=f(x_0)a=f(x_0)$$ 
as desired.
A: Your remark shows that $f(0)=1$ if $f$ is not trivial.
Let $x\in R$ you have to show that $lim_nf(x_n)=f(x)$ if the sequence $(x_n)$ converges towards $x$. You have $f(x_n)=f(x_n-x+x)=f(x_n-x)f(x)$ since $(x_n-x)$ converges towards $0$ $lim_nf(x_n-x)=f(0)=1$. This implies that $lim_nf(x)=lim_nf(x_n-x)f(x)=f(x)$.
A: Let $a \in \mathbb{R}$.First consider the case $f(a)=0$. Then, for all $x$, $f(x)=f(x-a+a)=f(a)f(x-a)=0$, so $f$ is continuous. Let $\epsilon>0$, and $f(a) \neq 0$. We want to show there exists $\delta>0$ such that whenever $|x-a|<\delta$, $|f(x)-f(a)|<\epsilon$.  Choose $\delta>0$ such that $$|f(h)-1|<\frac{\epsilon}{f(a)}$$ whenever $|h|<\delta$(here I use continuity in $0$). Then, if $x$ is such that $|x-a|<\delta$, we have
$$|f(x)-f(a)|=|f(a+x-a)-f(a)|=|f(a)f(x-a)-f(a)|=|f(a)||f(x-a)-1|<\epsilon.$$
Note: of course, this is no how I came up with this. How one comes up with something like this is noticing that for $h$ small enough: $|f(a+h)-f(a)|=|f(a)||f(h)-1|$, which is small because $f$ is continuous at $0$.
