## swissgirl 2 years ago Assume $$f: \mathbb{R} \to \mathbb{R}$$ is such that $$f(x+y)=f(x)f(y)$$ ( The class of exponential functions has this property). Prove that f having a limit at 0 implies that f has a limit at every real number and is one, or f is identically 0 for every $$x \in \mathbb{R}$$

1. satellite73

if $$f$$ is identically zero there is nothing to prove

2. swissgirl

Well you are proving backwards you gotta start from the beginning

3. satellite73

i guess we have to worry about $$f$$ being continuous start by showing $$f(0)=1$$

4. satellite73

this line Prove that f having a limit at 0 implies that f has a limit at every real number and is one, does not really make sense to me we can show that $$f(0)=1$$ but for example if $$f(x)=e^x$$ then what does "$$f$$ has a limit at every real number and is one" mean?

5. swissgirl

Not exactly sure hmmmmm

6. satellite73

$f(x)=f(x+0)=f(x)f(0)\implies f(0)=1$

7. swissgirl

Aha I see that

8. satellite73

if $$f$$ is continuous at $$0$$ then $\lim_{h\to 0}f(x+h)=\lim_{h\to 0}f(x)f(h)=f(x)$ making $$f$$ continuous

9. swissgirl

Thanks :) Just gonna string some stuff together here :)

10. joemath314159

you could write the hypothesis as a limit, then do a change of variable. Since the limit as x goes to 0 exists, we have:$\lim_{x\rightarrow 0}f(x)=c$for some c. Let b be a fixed real number, and let x = y-b. Then saying x goes to zero is equivalent to saying y goes to b. So we get this now:$\lim_{y\rightarrow b}f(y-b)=c\Longrightarrow \lim_{y\rightarrow b}f(y)(f(-b)=c$$\Longrightarrow f(-b)\lim_{y\rightarrow b}f(y)=c$Note that since f(0)=1, it follows that:$1=f(0)=f(x+(-x))=f(x)f(-x)$for all x. Therefore$f(-b)\lim_{y\rightarrow b}f(y)=c\Longrightarrow \lim_{y\rightarrow b}f(y)=cf(b)$So the limit exists at b. Since b was arbitrary, the limit exists everywhere.

11. swissgirl

Ohhh I like thisssss :)