Is $\lim\limits_{x \to c} f(x) = L$ equivalent to $\lim\limits_{x \to \frac{c}{2}} f(2x) = L$? This is a follow-up to a question I posted earlier today, whose answer I am still unsure of. I've been thinking about it some more, and I think I could answer it in a way that makes sense to me if the following claim is true:
$$\displaystyle\lim_{x \to c} f(x) = L \Longleftrightarrow \displaystyle\lim_{2x \to c} f(2x) = L\Longleftrightarrow \displaystyle\lim_{x \to \frac{c}{2}} f(2x) = L$$
I've just started to learn about limits, and I've been wrong often enough that I can't really be sure of anything. Could someone weigh in?
 A: Yes, the claim is true.
Formally, what we mean when we write things like $\lim\limits_{x \to c} f(x) = L$ is that for every $\varepsilon > 0$, there exists some $\delta > 0$ such that if $0 < |x - c| < \delta$, then $| f(x) - L | < \varepsilon$. (One might want to be a little bit careful with the domain of $f$ here, but let us not worry about that for now.)
Now if we replace $x$ by $2 x$ in the above, we haven't changed anything---you can think of it as $2 x$ simply being a new symbol, still representing a quantity that gets very close to the number $c$. 
We are then left with $0 < |2 x - c| < \delta$ and $|f(2x) - L| < \varepsilon$, for the same choices of $\varepsilon$ and $\delta$.
Finally, dividing the first inequality by $2$, we get $0 < |x - \frac{c}{2}| < \frac{\delta}{2}$. This doesn't change anything; it still describes the exact same inequality, just written differently, and $\frac{\delta}{2}$ is still a positive number.
Hence translating this back into limits, it reads 
$$\lim_{x \to \frac{c}{2}} f(2 x) = L.$$
