I'm trying to prove the following claim (conjecture).
Let $f\in C^{1}(0,1)$ be such that $f(\frac{1}{2})=0$ and $f>0$ in $(0,1)\setminus\{\frac{1}{2}\}$. Prove that $$ \lim_{x\to\frac{1}{2}}\frac{f(x)}{(\frac{1}{2}-x)^{a}}=0\quad\text{for any $a\in(0,1)$}. $$
My proof: From the assumption of $f$, the case $f\equiv const.$ is removable. In addition, the case $f=o((\frac{1}{2}-x)^{a})$ (this notation might be wrong) is also removable for any $a\in(0,1)$; otherwise, it contradicts to $f\in C^{1}$. Thus, since $f>o((\frac{1}{2}-x)^{a})$ for any $a\in(0,1)$, we have the conclusion.
I wonder if the proof in second and third sentences is not enough and there are more nice proofs. If you have some ideas, could you tell me that? And also, this statement might be wrong. If so, please point out which is wrong or give me a counterexample.
Thank you in advance.
$$ \lim_{x\to\frac{1}{2}}\frac{f(x)}{(\frac{1}{2}-x)^{a}}=\lim_{x\to\frac{1}{2}}\frac{f\left( x\right) -f\left( \frac{1}{2}\right) }{x-\frac{1}{2}} \cdot \lim_{x\to\frac{1}{2}} \left(-\left(\frac{1}{2} -x\right)^{1-a}\right) =f'\left(\frac{1}{2}\right)\cdot 0 =0$$