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 \begin{equation} \exists \delta > 0, \forall x, 0 < \lvert x-x_{0}\rvert < \delta \implies \lvert f\left(x\right) - L \rvert < \epsilon, \end{equation} 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 \begin{equation} \forall x, 0 < \lvert x-x_{0}\rvert < \delta \implies \lvert f\left(x\right) - L \rvert < g\left(\epsilon\right), \end{equation} 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?

  • $\begingroup$ 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. $\endgroup$ Jan 13, 2021 at 15:23
  • $\begingroup$ 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. $\endgroup$
    – Countable
    Jan 13, 2021 at 15:23
  • 4
    $\begingroup$ Note that in theses cases you always have $g(\varepsilon) \to 0$ as $\varepsilon \to 0$. Can you continue from here? $\endgroup$ Jan 13, 2021 at 15:23
  • $\begingroup$ @DaniloGregorinAfonso Right to the point, great! $\endgroup$
    – Ziqi Fan
    Jan 13, 2021 at 15:24

3 Answers 3


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 $|A|<g(\epsilon)$ for all $\epsilon>0$, where $\displaystyle\lim_{y\to 0+}g(y)=0$, you can conclude that $|A|=0$.

Suppose you have shown that $$ \forall \epsilon\ \exists \delta\ \forall x \ \ |x-x_0|<\delta\Rightarrow |f(x)-L|<g(\epsilon)\tag{1} $$ where $\displaystyle \lim_{y\to 0+}g(y)=0$, you can then show that $$ \forall \epsilon\ \exists \delta\ \forall x \ \ |x-x_0|<\delta\Rightarrow |f(x)-L|<\epsilon\tag{2} $$


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

$$ |x-x_0|<\delta\Rightarrow |f(x)-L|<g(r)<\epsilon. $$

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.


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