Vtyjkyuk

Let function $f: mathbbR to mathbbR$ periodic with period $T>0$. I want to show, using the definition of the limit at infinity, that $f$ cannot have the limit $infty$ while $x to infty$.

The definition is the following:

$lim_xto +infty f(x)=L$ means that for each $epsilon>0$ we can find a $N$ such that if $x>N$ then $|f(x)-L|<epsilon$.

Let $epsilon>0$ and $N$ such that $x>N$.

Since $f$ is periodic, if $|f(x)-L|< epsilon$ then $|f(x+T)-L|< epsilon$.

Then we have that $L-epsilon<f(x)<L+epsilon$ and $L-epsilon<f(x+T)<L+epsilon$.

Is the above procedure right so far? How can we continue?

No, it is not correct. You are aiming at proving that the limit of $f$ at $infty$, if it exists, is not $infty$. But then you start writing about the definition of $lim_xtoinftyf(x)=L$, where $L$ is a real number.

Suppose that $lim_xtoinftyf(x)=infty$. Then there is a $M>0$ such that $x>Mimpliesbigllvert f(x)bigrrvert>bigllvert f(0)bigrrvert$. But that's impossible, because$$f(0)=f(T)=f(2T)=cdots$$and, if $Ninmathbb N$ is large enough, then $nT>M$ and therefore $bigllvert f(nT)bigrrvert$ should be greater than $bigllvert f(0)bigrrvert$.

$f(nT)=f(0)$ for all positive integers $n$. If $f(x) to infty$ as $ x to infty$ then we get $f(0)=infty$ contradicting the fact that $f$ is real valued. Incidentally, you have written the definition of $f$ having a fin ite limit. For $f(x) to infty$ the definition is: given any positive number $A$ there exists $t$ such that $x >t$ implies $f(x) >A$. Show that this cannot happen by taking $A>f(0)$ and $x=nT$ with $n >frac t T$.

You may reason as follows:

As $f$ is not constant there is a $xi in [0, T)$ such that $f(xi) neq f(0)$.

Now, set $x_n := nT$ and $xi_n = xi+nT$. So, you get
$$lim_nrightarrow infty f(x_n) = f(0) mbox and lim_nrightarrow infty f(xi_n) = f(xi)$$

Consequently, $lim_xrightarrow infty f(x)$ does not exist and in addition

$$f(x) stackrelxrightarrow inftynot rightarrow infty,$$

as $(f(x_n))_n in mathbbN$ and $(f(xi_n))_n in mathbbN$ are bounded.