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?
Best Answer
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).