Suppose $f$ is a bounded function on $[0,1]$ satisfying $f(ax) = bf(x)$ for $0 \leq x \leq \frac{1}{a}$ and $a,b > 1.$ Prove that $\lim_{x \to 0^+} f(x) = f(0).$
Proof (attempt)
Notice that $f( a (ax) ) = b f(ax) = b^2 f(x) $ and in generalt $f(a^n x) = b^n f(x) $ for $0 \leq x \leq a^{-n} $
Let $\epsilon > 0$ be given and we need to find $\delta > 0$ such that if $x < \delta$, then $|f(x) - f(0) | < \epsilon $ or that $|f(x)| < \epsilon $ because since $f(0) = b f(0) $ and $b > 1$ ,then $f(0) = 0$.
We know that $f(x)$ is bounded so $|f(x)| < M $ for some $M > 0$ and $x \in [0,1]$. In particular $|f(a^n x)| < M $. Therefore, we have that
$$ |f(x) | < \dfrac{M}{b^n} $$
Now, $Mb^{-n} < \epsilon $ iff $n > \dfrac{ \ln (M / \epsilon) }{\ln b } $. So, if we choose $\delta > \dfrac{1}{a^n} $ for $n > \dfrac{ \ln (M / \epsilon) }{\ln b } $ then $|f(x)-f(0)| < \epsilon $ as was to be proved.
Is this correct proof? I am still not quite sure if my choice of $\delta > 0$ is the correct one.
You have to be more precise on the order of definition of quantities.
1-Let be $\epsilon$
2-Let be $N$
3- Let be $\delta$.
Precise it (e.g. $N= \left \lceil{\dfrac{\ln(M/\epsilon}{\ln(b)}}\right \rceil$), where $\left \lceil{x}\right \rceil$ denotes the smallest integer greater than your quantity. It is actually more rigourous than just giving an inequality if you have to give a piece of work.
Then you can do your work properly with fixed $N$.
Otherwise your work is correct.