Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

swissgirl

  • 3 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 \( x \in \mathbb{R}\)

  • This Question is Closed
  1. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    if \(f\) is identically zero there is nothing to prove

  2. swissgirl
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Well you are proving backwards you gotta start from the beginning

  3. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  4. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 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?

  5. swissgirl
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Not exactly sure hmmmmm

  6. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[f(x)=f(x+0)=f(x)f(0)\implies f(0)=1\]

  7. swissgirl
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Aha I see that

  8. anonymous
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 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

  9. swissgirl
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  10. joemath314159
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 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 = 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
    • 3 years ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ohhh I like thisssss :)

  12. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy