How can I prove $\lim_{x \to a} f(x)=L$ implies $\lim_{x \to 3a}f(x/3)=L$ with the epsilon-delta definition of the limit?

Question

How can I prove $\lim_{x \to a} f(x)=L$ implies $\lim_{x \to 3a}f(x/3)=L$ with the epsilon-delta definition of the limit?

I have tried doing this by writing down separately the epsilon-delta definitions of $\lim f(x)=L$ as $x$ approaches $a$ and $\lim f(x/3)=L$ as $x$ approaches $3a$. I have no idea how to proceed from here.

Could anyone please give me a tip? This is a homework problem.

Answer 1

Assume, we have $\forall\varepsilon, \exists \delta, \forall x $ $$0<|x-a|< \delta \Rightarrow |f(x)-L|<\varepsilon$$ Let's consider $x=\frac{t}{3}$ one-to-one map $\mathbb{R}\to\mathbb{R}$.

Using it set $0<|x-a|=\left|\frac{t}{3}-a\right| = \frac{1}{3}\left|t-3a\right|< \delta$ transformed to set $0<|t-3a|<3 \delta$, so taking $\delta=\delta_1/3$ gives $$0<|t-3a|< \delta_1 \Rightarrow \left|f\left(\frac{t}{3}\right)-L\right|<\varepsilon$$

Answer 2

Epsilon delta proofs are all about finding a delta based on epsilon so that x chosen in that delta interval around your point works. Another useful piece of advise is to write out what you can assume definitionally with these kinds of "by the definition" proofs and write out what you want to prove as well, in an actual mathematical statement. So if we want f(x) to be epsilon close to L, we have a delta so that if $x\in (a-\delta,a+\delta)$, then $\lvert f(x)-L\rvert <\varepsilon$. We want to find a delta (which I'll denote as $\gamma$ for clarity) so that $x\in (x-\gamma, x+\gamma)\implies \lvert f(\frac{x}{3})-L\rvert <\varepsilon$. I can take the same epsilon since the first inequality is true for all epsilon, and I want to prove the second also holds for any epsilon.

We know that $\frac{x}{3}\in (a-\delta,a+\delta)$, then $\lvert f(\frac{x}{3})-L\rvert <\varepsilon$, but this is basically exactly what we want!. So we just have to manipulate a little, i.e., $\lvert \frac{x}{3}-a\rvert <\delta \iff \lvert x -3a\rvert <3\delta$. So if we set $\gamma$ to be $3\delta$, then we have our proof!