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 one year 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 \( x \in \mathbb{R}\)
 one year 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 \( x \in \mathbb{R}\)

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

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

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0this 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?

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

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

satellite73
 one year ago
Best ResponseYou've already chosen the best response.0if \(f\) 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

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

joemath314159
 one year ago
Best ResponseYou've already chosen the best response.2you 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.

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