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swissgirl
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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 \( x \in \mathbb{R}\)
 one year ago
 one year ago
swissgirl Group Title
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 \( x \in \mathbb{R}\)
 one year ago
 one year ago

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satellite73 Group TitleBest ResponseYou've already chosen the best response.0
if \(f\) is identically zero there is nothing to prove
 one year ago

swissgirl Group TitleBest ResponseYou've already chosen the best response.0
Well you are proving backwards you gotta start from the beginning
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
i guess we have to worry about \(f\) being continuous start by showing \(f(0)=1\)
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
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?
 one year ago

swissgirl Group TitleBest ResponseYou've already chosen the best response.0
Not exactly sure hmmmmm
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
\[f(x)=f(x+0)=f(x)f(0)\implies f(0)=1\]
 one year ago

swissgirl Group TitleBest ResponseYou've already chosen the best response.0
Aha I see that
 one year ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

swissgirl Group TitleBest ResponseYou've already chosen the best response.0
Thanks :) Just gonna string some stuff together here :)
 one year ago

joemath314159 Group TitleBest ResponseYou've already chosen the best response.2
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 = yb. Then saying x goes to zero is equivalent to saying y goes to b. So we get this now:\[\lim_{y\rightarrow b}f(yb)=c\Longrightarrow \lim_{y\rightarrow b}f(y)(f(b)=c\]\[\Longrightarrow f(b)\lim_{y\rightarrow b}f(y)=c\]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.
 one year ago

swissgirl Group TitleBest ResponseYou've already chosen the best response.0
Ohhh I like thisssss :)
 one year ago
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