# 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!

• One idea would be to find the various values of $\delta$ for the first limit, at $x, x+1, x+2, x+3$, that yield a deviation from $1$ of no more than $\sqrt[4]{1+\varepsilon}$, or something like that. Then when you multiply all those together, you stay within $\varepsilon$ of $1$. I may have gotten some of the adjustments wrong, but that's one basic idea. Dec 20, 2021 at 21:00
• Oops, I guess rather than $\delta$, that would be $N$ (or whatever you want to call it). But same basic idea. Dec 20, 2021 at 21:05
• @BrianTung Uh, it doesn't really matter. The definitions do not depend on the choice of the letter anyway. Interesting idea, but I'm not exactly sure how to make it work. I am too tired now to continue for today. Dec 20, 2021 at 21:08
• @LorentzianMcDonalds, I am not sure that the $\epsilon-\delta$ can be applied here because the function is arbitrary. Dec 20, 2021 at 21:17
• @LorentzianMcDonalds: I'm aware that the letter doesn't matter in any precise way, but $\delta$ is usually used to denote the half-width of the interval around $x$, whereas $N$ (or something similar) is used to denote a value of $x$ beyond which $f(x)$ is always within $\varepsilon$ of $1$ (in this case). In other words, these are conventions which make reading mathematics easier, in much the same way we usually use $n$ to refer to an integer, $x$ to a real value, $z$ to a complex value, and so on. Dec 20, 2021 at 21:47

## 3 Answers

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 $$1-\epsilon<{\frac{f(x+1)}{f(x)}\cdot\frac{f(x+2)}{f(x+1)}\cdot\frac{f(x+3)}{f(x+2)}\cdot\frac{f(x+4)}{f(x+3)}}<1+\epsilon.$$

• I have a question. How do you handle $(1-\epsilon)^{1/4}$ for $\epsilon > 1$? Dec 21, 2021 at 0:21
• ^This is just book-keeping. Rewrite the proof with $\epsilon' = \min(1,\epsilon),$ then if you find an $M$ such that after it the function is bounded within $1 \pm \epsilon',$ then it is a fortiori bounded within $1 \pm \epsilon$. Dec 21, 2021 at 2:21

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 $$h$$.

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.