Show that limit cannot be $\infty$

Question

Let function $f: \mathbb{R} \to \mathbb{R}$ 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:

$\lim_{x\to +\infty} f(x)=L$ 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?

Answer 1

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 $\lim_{x\to\infty}f(x)=L$, where $L$ is a real number.

Suppose that $\lim_{x\to\infty}f(x)=\infty$. Then there is a $M>0$ such that $x>M\implies\bigl\lvert f(x)\bigr\rvert>\bigl\lvert f(0)\bigr\rvert$. But that's impossible, because$$f(0)=f(T)=f(2T)=\cdots$$and, if $N\in\mathbb N$ is large enough, then $nT>M$ and therefore $\bigl\lvert f(nT)\bigr\rvert$ should be greater than $\bigl\lvert f(0)\bigr\rvert$.

Answer 2

$f(nT)=f(0)$ 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 $n >\frac t T$.

Answer 3

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 $$\lim_{n\rightarrow \infty} f(x_n) = f(0) \mbox{ and } \lim_{n\rightarrow \infty} f(\xi_n) = f(\xi)$$

Consequently, $\lim_{x\rightarrow \infty} f(x)$ does not exist and in addition

$$f(x) \stackrel{x\rightarrow \infty}{\not \rightarrow} \infty,$$

as $(f(x_n))_{n \in \mathbb{N}}$ and $(f(\xi_n))_{n \in \mathbb{N}}$ are bounded.