swissgirl
  • 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 \( x \in \mathbb{R}\)
Mathematics
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
if \(f\) is identically zero there is nothing to prove
swissgirl
  • swissgirl
Well you are proving backwards you gotta start from the beginning
anonymous
  • anonymous
i guess we have to worry about \(f\) being continuous start by showing \(f(0)=1\)

Looking for something else?

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

More answers

anonymous
  • anonymous
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?
swissgirl
  • swissgirl
Not exactly sure hmmmmm
anonymous
  • anonymous
\[f(x)=f(x+0)=f(x)f(0)\implies f(0)=1\]
swissgirl
  • swissgirl
Aha I see that
anonymous
  • anonymous
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
swissgirl
  • swissgirl
Thanks :) Just gonna string some stuff together here :)
anonymous
  • anonymous
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.
swissgirl
  • swissgirl
Ohhh I like thisssss :)

Looking for something else?

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