How to prove $\lim_{x→a/2} f(2x) = L$ if we know that $\lim_{x→a} f(x) = L$.

Question

Let $f$ be a function with domain $R$. Let $a, L ∈ R$.

Assume that $\lim_{x→a} f(x) = L$

how do we prove using delta epsilon that $\lim_{x→a/2} f(2x) = L$ ?

Answer 1

$\forall \epsilon>0,\exists \delta\mid |u-a|<\delta\implies |f(u)-L|<\epsilon$

Now for the same epsilon, same delta and $u=2x$ we have

$|2x-a|<\delta\implies|f(2x)-L|<\epsilon$

Which can be rewritten $|x-\frac a2|<\frac{\delta}2\implies|f(2x)-L|<\epsilon$

So $\forall \epsilon>0,\exists \gamma=\frac{\delta}2\mid |x-\frac a2|<\gamma\implies |f(2x)-L|<\epsilon$

First line and last line are exactly the definitions of the limits given in the wording of the problem.

Answer 2

$$\lim_{x\to a}f(x) \equiv f(a)$$

and $$\lim_{x\to \frac{a}{2}} f(2x)=f(2\times\frac{a}{2})\equiv f(a)$$

Need more explanation?

Edit: Assumptions- The function $f(x)$ is a continous real function defined in $x\in R$. So, I assume that $f(a)$ exists