# Thoughts about the use of $\epsilon$-$\delta$ language in proofs involving limits

The limit of a function $$f: \mathbb{R} \to \mathbb{R}$$ at $$x_{0}$$ is defined as follows:

$$\lim_{x \to x_{0}} f\left(x\right) = L \iff \forall \epsilon > 0, \exists \delta > 0, \forall x, 0 < \lvert x-x_{0}\rvert < \delta \implies \lvert f\left(x\right) - L \rvert < \epsilon$$.

In proofs of limits using the definition, we typically fix an arbitrary constant $$\epsilon > 0$$, and as long as we are able to prove $$$$\exists \delta > 0, \forall x, 0 < \lvert x-x_{0}\rvert < \delta \implies \lvert f\left(x\right) - L \rvert < \epsilon,$$$$ we can conclude that $$\lim_{x\to x_{0}}f\left(x\right) = L$$. The key is to construct such a $$\delta$$ so that the above relation holds. However, as I saw in some textbooks, for an arbitrary $$\epsilon > 0$$, if we are able to find a $$\delta > 0$$ such that $$$$\forall x, 0 < \lvert x-x_{0}\rvert < \delta \implies \lvert f\left(x\right) - L \rvert < g\left(\epsilon\right),$$$$ where $$g\left(\epsilon\right)$$ is a term containing $$\epsilon$$ ($$g\left(\epsilon\right) = 3\cdot\epsilon$$, for example), then we can still claim that $$\lim_{x\to x_{0}}f\left(x\right) = L$$.

I am wondering if there exists a simple proof for using the general $$g\left(\epsilon\right)$$ in proofs of limits?

• Welcome to Math.SE! I have tried to improve the readability of your question by improving the $\rm \LaTeX$ code. It is possible that I unintentionally changed the meaning of your question. Please proofread the question to ensure this has not happened. Jan 13, 2021 at 15:23
• If g some sort of function of the form $ax^n$ where a is a real number then probably yes, but if a is the gamma function or say $g(x) = \frac{1}{x}$ probably not. Jan 13, 2021 at 15:23
• Note that in theses cases you always have $g(\varepsilon) \to 0$ as $\varepsilon \to 0$. Can you continue from here? Jan 13, 2021 at 15:23
• @DaniloGregorinAfonso Right to the point, great! Jan 13, 2021 at 15:24

If you have $$|A|<3\epsilon$$ for any $$\epsilon>0$$, you can conclude that $$|A|<\epsilon$$ for any $$\epsilon>0$$, which implies that $$|A|=0$$.

In general, if you have the estimates of the form $$|A| for all $$\epsilon>0$$, where $$\displaystyle\lim_{y\to 0+}g(y)=0$$, you can conclude that $$|A|=0$$.

Suppose you have shown that $$\forall \epsilon\ \exists \delta\ \forall x \ \ |x-x_0|<\delta\Rightarrow |f(x)-L| where $$\displaystyle \lim_{y\to 0+}g(y)=0$$, you can then show that $$\forall \epsilon\ \exists \delta\ \forall x \ \ |x-x_0|<\delta\Rightarrow |f(x)-L|<\epsilon\tag{2}$$

Proof.

Let $$\epsilon>0$$. Since $$\displaystyle \lim_{y\to 0+}g(y)=0$$, there exists $$r>0$$ such that $$g(r)<\epsilon$$. By (1), there exists $$\delta>0$$ such that

$$|x-x_0|<\delta\Rightarrow |f(x)-L|

Now you have (2).

As long as $$g(\epsilon)$$ goes infinitely close to $$0$$ while staying positive, as $$\epsilon \to 0$$, i.e. $$\lim_{\epsilon\to 0} g(\epsilon) = 0$$ and $$g(\epsilon) > 0$$, any such $$g$$ will do because

$$\forall \zeta > 0,\ \exists \gamma > 0, \ 0 < \epsilon < \gamma \implies g(\epsilon) < \zeta$$

So, if using an $$\epsilon-\delta$$ technique we conclude that $$\exists$$ a suitable $$\delta$$ such that $$|f(x) - L| < g(\epsilon)$$, then $$\forall \zeta > 0$$, we choose $$\epsilon < \gamma$$ and $$|f(x) - L| < g(\epsilon) < \zeta$$ whenever $$0 < |x-x_0| < \delta$$.

To formalize @DaniloGregorinAfonso's idea, you may set $$\epsilon' = g(\epsilon)$$, so as to cheer up yourself for the moment, giving a hope that you can "clean up the RHS". $$\forall \epsilon > 0, \exists \delta > 0, \forall x, 0 < \lvert x-x_{0}\rvert < \delta \implies \lvert f\left(x\right) - L \rvert < \epsilon'$$ Assume that $$\epsilon = g^{-1}(\epsilon')$$. Note that $$\delta$$ is a function $$\epsilon$$, so you may write $$\delta = \delta(\epsilon) = \delta(g^{-1}(\epsilon'))$$. You may find the notation $$g^{-1}$$ problematic, say, in case that $$g(\epsilon) = \begin{cases}\epsilon (\sin(1/\epsilon) + 2021/2020) & \epsilon \ne 0 \\ 0 & \epsilon = 0. \end{cases}$$ In such case, you may find a monotone increasing upper bound $$h(\epsilon) \ge g(\epsilon)$$ such that $$h(\epsilon) \to 0$$ as $$\epsilon \to 0$$, and do that same thing to get $$\delta(\epsilon) = \delta(h^{-1}(\epsilon'))$$. Now you get a function $$\delta$$ which depends on the new $$\epsilon'$$. $$\forall \epsilon > 0, \exists \delta(h^{-1}(\epsilon')) > 0, \forall x, 0 < \lvert x-x_{0}\rvert < \delta \implies \lvert f\left(x\right) - L \rvert < \epsilon'$$ Finally replace $$\epsilon$$ in "$$\forall \epsilon > 0$$" on the left-hand side by $$\epsilon'$$.

It remains to show that such monotone increasing upper bound $$h$$ for $$g$$ exists, but we may construct it like $$h(\epsilon) = \sup\limits_{t \in (0,\epsilon)} |g(\epsilon)|$$. It's clear that $$h$$ is monotone increasing. It's easy to verify that $$h\searrow0$$ as $$\epsilon \searrow 0$$.

$$\forall \epsilon > 0, \exists \delta > 0, \forall t \in (0,\delta), |g(t)| < \epsilon \\ \implies \sup_{t \in (0,\delta)} |g(t)| < \epsilon \\ \implies \forall t \in (0,\delta), h(t) = \sup_{s \in (0,t)} |g(s)| \le \sup_{s \in (0,\delta)} |g(s)| < \epsilon$$

Add an extra term $$\epsilon$$ to $$h$$ guarantees that $$h$$ to be strictly increasing, i.e. the existence of $$h^{-1}$$, completing the proof.