Suppose that $f$ is a real-valued function defined on all of $R$ and satisfying the identity $$ f(x+y)=f(x) f(y)$$ for all $x, y$ in $\mathbb{R}$.
Prove that, if $f$ is continuous at $x=0$, then $f$ is continuous on all of $R$.
Initially, one has to prove that $f(0)=1$ since otherwise $f$ would be identically equal to $0$, as well as $f(x) \geq 0$ is true for all $x \in R$. Moreover, the main approach seems to use the epsilon-delta technique, however, I couldn’t be able to prove mathematically how this relation can be correct using the above identity.
Fix $x\in\Bbb R\setminus\{0\}$. Then for every $y\in\Bbb R$ it holds that $$f(x+y)-f(x)=f(x)f(y)-f(x)=f(x)(f(y)-1)=f(x)(f(y)-f(0))$$
Assume $f$ is contoinuous at $0$, so $\lim_{y\to0}f(y)=f(0)$. Then $$\lim_{y\to0}(f(x+y)-f(x))=\lim_{y\to0}f(x)(f(y)-f(0))=0$$ and this menas that $$\lim_{y\to0}f(x+y)=f(x)$$
So $f$ is continuous at $x$.