Paraphrasing the delta-epsilon definition of limits on Wikipedia,
"Given function $f:\mathbb{R}\to\mathbb{R}$ and reals $a,L\in\mathbb{R}$. We say $\stackrel{\lim}{_{x\to a}} f(x)=L$ if $P(L)$ holds"
Where
$P(L):=\left[\forall \varepsilon\in \mathbb{R}_{>0}:\exists \delta\in \mathbb{R}_{>0}:\forall x\in \mathbb{R}: (0<|x-a|<\delta)\implies \left(\left|f(x)-L\right|<\varepsilon\right)\right]$
In short, it says: $P(L)\implies \stackrel{\lim}{_{x\to a}} f(x)=L\hspace{2em}$ (1)
It doesn't say: $P(L)\iff \stackrel{\lim}{_{x\to a}} f(x)=L\hspace{2.6em}$ (2)
It seems that (1) is weaker than (2). Statement (1) suffices when we wish to prove $\stackrel{\lim}{_{x\to 3}}2x=6$ (simply show that $P(6)$ holds with $f(x)=2x, a=3$), but what if we want to prove the following?
"Assume $\stackrel{\lim}{_{x\to a}}f(x)$ and $\stackrel{\lim}{_{x\to a}}g(x)$ exist. It follows that $\stackrel{\lim}{_{x\to a}}\left( f(x)\cdot g(x) \right)=\left( \stackrel{\lim}{_{x\to a}}f(x) \right)\cdot \left( \stackrel{\lim}{_{x\to a}}g(x) \right)$"
Although I think the above statement is true, I don't think you can prove it given the delta-epsilon definition of limits described in (1). One can show that if $[\forall \varepsilon\in \mathbb{R}_{>0}:\exists \delta_1\in \mathbb{R}_{>0}: \forall x\in \mathbb{R}: (0<|x-a|<\delta_1)\implies (|f(x)-L|<\varepsilon)]$ holds and $[\forall \varepsilon\in \mathbb{R}_{>0}:\exists \delta_2\in \mathbb{R}_{>0}: \forall x\in \mathbb{R}: (0<|x-a|<\delta_2)\implies (|g(x)-M|<\varepsilon)]$ holds for some $L,M\in \mathbb{R}$, then it follows that $\stackrel{\lim}{_{x\to a}}\left( f(x)\cdot g(x) \right)=\left( \stackrel{\lim}{_{x\to a}}f(x) \right)\cdot \left( \stackrel{\lim}{_{x\to a}}g(x) \right)=L\cdot M$
The issue is that given $\stackrel{\lim}{_{x\to a}} f(x)$ exists, (let's call it $L$) it doesn't follow that $[\forall \varepsilon\in \mathbb{R}_{>0}:\exists \delta_1\in \mathbb{R}_{>0}: \forall x\in \mathbb{R}: (0<|x-a|<\delta_1)\implies (|f(x)-L|<\varepsilon)]$
Likewise, given $\stackrel{\lim}{_{x\to a}} g(x)$ exists, (let's call it $M$) it doesn't follow that $[\forall \varepsilon\in \mathbb{R}_{>0}:\exists \delta_2\in \mathbb{R}_{>0}: \forall x\in \mathbb{R}: (0<|x-a|<\delta_2)\implies (|g(x)-M|<\varepsilon)]$
I suppose a more general question is, whenever a math text says something like "We say $x$ is [terminology] if $x$ has [property]" e.g. "We say a natural $x$ is happy if $x$ is divisible by 7", am I supposed to interpret this text as meaning "We say a natural $x$ is happy if and only if $x$ is divisible by 7"? The first statement says that 0, 7, 14 etc are happy numbers, but says nothing about whether or not 3 is happy, while the second says 3 is definitively not a happy number.
