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This question already has an answer here:

let $f$ Periodic function (Except Constant functions) then :

$\lim_{x\to \infty} f(x)= \text{Does not exist}$

is it right ??

such that :

$$\lim_{x\to \infty} \sin x= \text{Does not exist}$$

$$\lim_{x\to \infty} \cos x= \text{Does not exist}$$

$$\lim_{x\to \infty} \tan x= \text{Does not exist}$$

$$\lim_{x\to \infty} \cot x= \text{Does not exist}$$

$$\lim_{x\to \infty} \lfloor x\rfloor +\lfloor -x \rfloor= \text{Does not exist}$$

is it right ??

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marked as duplicate by Did, Rohan, Simply Beautiful Art, Daniel Fischer Jan 22 '17 at 12:14

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    $\begingroup$ Constant functions? $\endgroup$ – David Mitra Jan 22 '17 at 12:08
  • $\begingroup$ If, however, you consider only periodic functions where a primitive period exists, then the conjecture is correct. $\endgroup$ – Git Gud Jan 22 '17 at 12:15
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This is wrong. Every constant function is periodic but $\lim_{x \to \infty} f(x)$ exists. Furthermore it is difficult to consider something like $\lim_{x \to \infty} \tan(x)$.

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    $\begingroup$ It is meaningless, even, to consider the limit of $\tan$ at infinity because for the definition of limit at infinity it is required that an interval $]a,+\infty[$ for some $a$ is contained in the domain of the function. $\endgroup$ – Git Gud Jan 22 '17 at 12:14
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Wrong. There does exist a periodic function with a limit:

$$f(x)=c$$

for a constant $c$. Then the limit is trivial, but without assuming this to be the only case, you will find:

for $\lim_{x\to\infty}f(x)$ to exist, and $f(x)=f(x+T)$, then it must be the case that for all $a,b\in[0,T)$, then $f(a)=f(b)$, so it must be a constant function.

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