Proof that $\lim_{x\rightarrow \infty} f(x+a) =\lim_{x\rightarrow \infty} f(x) $ for constant $a$ It is easy to understand why the following statement is true for variable $x$ and constant $a$:
$\lim_{x\rightarrow \infty} f(x+a) =\lim_{x\rightarrow \infty} f(x) $
However, is there a proof for it, or is it true by definition?
 A: You're almost there. Let $l:= \lim\limits_{x\to \infty} f(x)$. Now, let $\epsilon>0$ be arbitrary. Then, by hypothesis, there exists a $c\in \Bbb{R}$ such that for all $x\in \Bbb{R}$, if $x>c$ then $|f(x)-l| < \epsilon$.
I'm not sure if this will help you or not, but let's just temporarily define $g(x):= f(x+a)$. Our goal is to show $\lim\limits_{x\to \infty}g(x) = l$ as well. Now, note that
\begin{align}
x>c-a & \implies x+a > c \\
&\implies |f(x+a) - l| < \epsilon \\
& \implies |g(x) - l| < \epsilon
\end{align}
So, what we have done is shown the existence of a $\tilde{c}$ (namely $c-a$) such that if $x>\tilde{c}$ then $|g(x)-l| < \epsilon$. Therefore, we have shown that $\lim\limits_{x\to \infty}g(x) = l$, exactly as we wanted to.
A: Yes, you can write $y = x + a$.
We can see that $x \rightarrow \infty$ is equivalent to $y \rightarrow \infty$ and so $\lim_{x\rightarrow \infty} f(x+a) = \lim_{y \rightarrow \infty}f(y) = \lim_{x\rightarrow \infty} f(x) $
If you want to show it using the definition of limit:
The limit of $f(x)$, as $x \rightarrow \infty$ is L, if for every $\epsilon \gt 0$ there is a number $N$ such that $\forall x > N\ ;|f(x) − L| \lt \epsilon$.
This means, in particular, that $\forall x > N - a$, we get that $ x + a > N $ and $|f(x + a) − L| \lt \epsilon$, so L is also the limit of $f(x + a)$ at infinity.
