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Let $f: \mathbb R \to \mathbb R$ fulfill the following $f(x+1)=f(x)$.

1.) Prove that for all $n \in Z$ $f(x+n)=f(x)$.

2.) Prove that if the limit $\lim_{x \to \infty} f(x)=L$ exists, then f is a constant.

1.) Since we're given that $n \in Z$, I was immediately drawn to using induction. f is a periodic function with a period of 1.

Let's try n=1:

$f(x+1)=f(x)$

This is correct because that's what was given. Let's assume for any $n\in N$ this statement still holds true. Let's try for $n+1$:

$f(x+(n+1)$

Since we assumed n is true we can set (n+1)=m, and $m\in N$,therefore

$f(x+m)=f(x)$

Is my work correct? By induction we assume it holds true for n, adding 1 is just adding another period to the function.

2.) I'm having difficulties even starting with this one. Was thinking about using derivatives but it's not explicitly given that f is differentiable.

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  • $\begingroup$ For question #1, what about non-positive $n$? $\endgroup$ – Kenny Lau Jul 30 '16 at 10:00
  • $\begingroup$ @KennyLau you're right. I'm so used to seeing n \in N that I overlooked an entire group of numbers/didn't read the question correctly. $\endgroup$ – RonaldB Jul 30 '16 at 10:04
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    $\begingroup$ For question #2, if $f$ is not a constant, then it oscillates and does not have a limit. $\endgroup$ – Kenny Lau Jul 30 '16 at 10:07
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Your proof is good for proving that $f(x+n)=f(x)$ for positive integers $m$.

However, if you notice that $$ f(x-1)=f((x-1)+1)=f(x) $$ you'll be able to complete the proof for negative integers as well.

For the second part, assume $f$ is not constant, so there are $x_1$ and $x_2$ with $f(x_1)\ne f(x_2)$. Consider the sequences $$ a_n=x_1+n,\qquad b_n=x_2+n $$ What can you say about $$ \lim_{n\to\infty}f(a_n) $$ and $$ \lim_{n\to\infty}f(b_n) $$ taking into account that $\lim_{x\to\infty}f(x)=L$?

If you can't use sequences, it's a bit more complicated. But you can notice that, given $M>0$, there always is $n$ such that $x_1+n>M$ and $x_2+n>M$.

Take $\varepsilon=|f(x_1)-f(x_2)|/2$ and try seeing what happens taking into account that $\lim_{x\to\infty}f(x)=L$.

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  • $\begingroup$ Ahh we weren't taught sequences/series in this course besides Taylor so won't be able to use that method, but thanks for helping me complete #1. $\endgroup$ – RonaldB Jul 30 '16 at 10:14
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    $\begingroup$ @RonaldB No problem: you can use them in disguise to make an $\varepsilon$-$\delta$ proof. $\endgroup$ – egreg Jul 30 '16 at 10:23
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$f$ is $1-$periodic. Therefore, if $f$ is not constant, there is $x, y$ s.t. $f(x)\neq f(y)$. In particular, $$f(x)=f(x+n)\neq f(y+n)=f(y)$$ for all $n$. By taking $n\to \infty $, you get that $$\lim_{n\to \infty }f(x+n)=f(x)\neq f(y)=\lim_{n\to \infty }f(y+n)$$ and thus $\lim_{x\to \infty }f(x)$ doesn't exist, which is a contradiction.

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  • $\begingroup$ $f(x)=\lim_{x\to \infty }f(x+n)$ must be $f(x)=\lim_{n\to \infty }f(x+n)$ $\endgroup$ – user261263 Jul 30 '16 at 10:10
  • $\begingroup$ This is not true. For instance, $1/n \neq 0$ for all $n$, although $\lim (1/n) = 0$. $\endgroup$ – user258700 Jul 30 '16 at 10:32
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    $\begingroup$ @AhmedHussein: This proof is correct. I construct two sequence $(a_n=x+n)_n$ and $(b_n=y+n)_n$ that diverge to $+\infty $ and s.t. $\lim_{n\to \infty }f(a_n)\neq \lim_{n\to \infty }f(b_n)$ $\endgroup$ – Surb Jul 30 '16 at 10:45
  • $\begingroup$ @Surb sorry, the way it was written at first seemed to imply otherwise. $\endgroup$ – user258700 Jul 30 '16 at 10:54

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