Epsilon-Delta proof of an infinite limit Whilst searching in some of my older books, I found this particular exercise:

If $\displaystyle\lim_{x \to +\infty}{\frac{f(x+1)}{f(x)}=1,}$ prove that$\displaystyle\lim_{x \to +\infty}{\frac{f(x+4)}{f(x)}=1.}$

Now, I managed to solve this problem with a trick by constantly applying the substitution $x\to x+1$. So we get:
$\displaystyle\lim_{x \to +\infty}{\frac{f(x+2)}{f(x+1)}=1,} \displaystyle\lim_{x \to +\infty}{\frac{f(x+3)}{f(x+2)}=1,}$ and $\displaystyle\lim_{x \to +\infty}{\frac{f(x+4)}{f(x+3)}=1.}$  In the end,
$\displaystyle\lim_{x \to +\infty}{\frac{f(x+1)}{f(x)}\cdot\frac{f(x+2)}{f(x+1)}\cdot\frac{f(x+3)}{f(x+2)}\cdot\frac{f(x+4)}{f(x+3)}=\displaystyle\lim_{x \to +\infty}{\frac{f(x+4)}{f(x)}}}=1$.
But I am looking for something closer to using the definition of $\epsilon-\delta$ and I was wondering if that truly was possible. I searched but was not able to find a solution. Thanks!
 A: This is sort of completing @Brian Tung's thought:
Indeed, the most natural way to see the limit is the way you did it:
$$
\displaystyle\lim_{x \to +\infty}{\frac{f(x+1)}{f(x)}\cdot\frac{f(x+2)}{f(x+1)}\cdot\frac{f(x+3)}{f(x+2)}\cdot\frac{f(x+4)}{f(x+3)}}=1
$$
Let $\epsilon>0$. By hypothesis, there exists $M>0$ large such that $(1-\epsilon)^\frac14<\frac{f(x+1)}{f(x)}<(1+\epsilon)^\frac14$ when $x>M$. Replacing $x$ with $x+k$ for $k=1,2,3$, we have
$$
(1-\epsilon)^\frac14<\frac{f(x+k+1)}{f(x+k)}<(1+\epsilon)^\frac14
$$
when $x+k>x>M$. Hence, when $x>M$, we have
$$1-\epsilon<{\frac{f(x+1)}{f(x)}\cdot\frac{f(x+2)}{f(x+1)}\cdot\frac{f(x+3)}{f(x+2)}\cdot\frac{f(x+4)}{f(x+3)}}<1+\epsilon.$$
A: Let $g(x) := f(x+1)/f(x)$. Then, $g(x+1)g(x) = f(x+2)/f(x)$.
We just consider only the case $f(x+2)/f(x)$, because your case can be easily shown by repeating a similar process.
Since $\lim_{x\to\infty}g(x) = 1$,
$$
\forall \epsilon > 0: \exists \delta > 0: \forall x > \delta:|g(x)-1| < \epsilon
$$
Let $\epsilon_1 > 0$ and $a > 0$. Then, there is $\delta_1 > 0$ such that $\forall x > \delta_1:|g(x)-1| < a\epsilon_1$.
Since $x+1 > x > \delta_1$, $|g(x+1) -1| < a\epsilon_1$.
On the other hand,
$$
(g(x+1) -1)(g(x) - 1)
=
g(x+1)g(x)-g(x+1)-g(x) + 1
$$
or equivalently,
$$
g(x+1)g(x) -1
=
(g(x+1)-1)(g(x)-1) +g(x+1) + g(x) - 2
$$
Therefore, for all $x > \delta_1$,
$$
\begin{aligned}
|g(x+1)g(x) -1|
&\le
|g(x+1)-1||g(x)-1| + |g(x+1) - 1| + |g(x) - 1|\\
&<
a^2\epsilon_1^2 + 2a\epsilon_1
\end{aligned}
$$
We can choose $a > 0$ such that $a^2\epsilon_1^2 + 2a\epsilon_1 \le \epsilon_1$ because the left-hand side is a continuous increasing function of $a$, which is zero at $a = 0$, and goes infinity as $a \to \infty$.
Therefore, we conclude that whenever $x > \delta_1$, $|g(x+1)g(x)-1| < \epsilon_1$. This shows $\lim_{x \to \infty} g(x+1)g(x) = 1$.
Let $h(x) = g(x+1)g(x)$ and do the same for $h$.
A: I just want to say that if you have a proof that the limit of a product is the product of the limits (when both exist), then that proof can be copy-pasted 3 times, each copy adjusted appropriately, such that you get a proof of your exercise. This is because your solution simply uses that product lemma 3 times. Thus any ε-δ proof of that lemma yields an ε-δ proof of your exercise. This is also known as proof unfolding. Of course, if you want something cleaner than just unfolded proofs, zugzug's approach works.
On the other hand, you should not be too keen on sticking to just the plain definitions and trying to avoid use of lemmas, because this is a recipe for long and ugly proofs. For example, suppose that you were asked to prove the following instead:
$
\def\lfrac#1#2{{\large\frac{#1}{#2}}}
$

If $\lim_{x→∞} \lfrac{f(x+8)}{f(x)} = 1$ and $\lim_{x→∞} \lfrac{f(x+5)}{f(x)} = 1$, then $\lim_{x→∞} \lfrac{f(x+1)}{f(x)} = 1$.

You would have a much harder time finding a clean proof that avoids anything that looks like the product lemma, and you would not really learn anything much from avoiding the lemma.
