3
$\begingroup$

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?

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 Jan 13 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 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? $\endgroup$ – Danilo Gregorin Afonso Jan 13 at 15:23
  • $\begingroup$ @DaniloGregorinAfonso Right to the point, great! $\endgroup$ – Ziqi Fan Jan 13 at 15:24
2
$\begingroup$

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 $$ \forall \epsilon\ \exists \delta\ \forall x \ \ |x-x_0|<\delta\Rightarrow |f(x)-L|<\epsilon\tag{2} $$

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

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

Now you have (2).

$\endgroup$
2
$\begingroup$

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 $0 < |x-x_0| < \delta$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.