Problem:If $f(x)\in C(0,1]$,and for every certain and fixed value $x_0\in(0,1]$,the limit $\lim_{n \to \infty}f(\frac{x_0}{2^n}) $ exists. Can we get the limit $\lim_{x \to 0^+}f(x)$ exists?
And I have proved the limit exists if $f(x)$ is uniformly continuous on $(0,1]$:
Brief proof with the additional condition:
Because of $\cup ^{\infty }_{n=1}(\frac{1}{2^{n}},\frac{1}{2^{n-1}} ]=(0,1]$,We can restrict our consideration to $x_0 \in(\frac{1}{2}, 1]$.
Then, $\forall \varepsilon >0$,for every certain and fixed $x_0\in(\frac{1}{2},1]$,
we suppose the corresponding sequence $a_n(x_0)=\frac{x_0}{2^n}$ converge to $A(x_0)(n \to \infty)$,
then,$\exists N(x_0)\in \mathbb{N_+} $,s.t.$\left | f(a_n(x_0))-A(x_0) \right | <\varepsilon $,when $n>N(x_0)$,
the uniform condition can get $A(x_0）\equiv A$（converges to the same value)
and restrict $N(x_0)$ that $\exists N$,s.t.,$N(x_0)<N,\forall x_0\in(\frac{1}{2},1]$
So when $n>N,\left| f(a_n(x_0))-A \right | <\varepsilon,\forall x_0\in(\frac{1}{2},1]$
and we supposed for the same $\epsilon$,$\exists \delta$,when $\left | x_1-x_2 \right | <\delta ,\left | f( x_1)-f(x_2) \right | <\epsilon $,
appropriately increase $N$,s.t.,$\frac{1}{2^N} <\delta $,then, we consider $\cup ^{\infty }_{n=N+1}(\frac{1}{2^{n-1}},\frac{1}{2^{n}}] =(0,\frac{1}{2^{N+1}} ]=\left (0,\delta _1 \right ] $
when $x\in(0,\delta _1)$,$\left | f(x)-A \right | \le \left | f(a_{2N}(x_0))-A \right | +\left | f(x)-f(a_{2N}(x_0)) \right |\le 2\varepsilon $
So, $\lim_{x \to 0^+}f(x)=A$
and I think the uniform condition is indispensible. However,I can't construct a specific counter example.Could you help me?Thanks!
