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swissgirl
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 \( x \in \mathbb{R}\)
if \(f\) is identically zero there is nothing to prove
Well you are proving backwards you gotta start from the beginning
i guess we have to worry about \(f\) being continuous start by showing \(f(0)=1\)
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?
Not exactly sure hmmmmm
\[f(x)=f(x+0)=f(x)f(0)\implies f(0)=1\]
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
Thanks :) Just gonna string some stuff together here :)
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.
Ohhh I like thisssss :)