Let $f:[0,1]\to \mathbb R$ be continuous. In addition $f$ is analytic on $[0,1)$. This implies that $f$ has countably many zeros on $[0,1)$ that can accumulate only at $1$. In addition every zero of $f$ in $[0,1)$ has finite multiplicity: For each $t\in [0,1)$ with $f(t)=0$ there is a smallest number $k=k(t)>0$ such that $f^{(k)}(t)\ne0$.
Can we say something about the multiplicity of the roots $t$ when $t$ approaches the boundary point $1$?
Does the order $k(t)$ stay bounded for $t\to 1$?
Consider $$g(z)=\prod_{n=1}^\infty\left(1-\frac{z^2}{\left\lceil \sqrt n\right\rceil^4}\right), $$ which is obtained from the product development of the sinc function by "pushing" some zeroes outward. The infinite product converges iff $\sum_{n=1}^\infty\frac{1}{\lceil \sqrt n\rceil^4}$ converges, and this follows from domimance by $\sum_{n=1}^\infty\frac{1}{n^2}$. Note that $$\underbrace{\left\lceil \sqrt{(m^2-2m+2}\right\rceil =\ldots =\left\lceil\sqrt{m^2}\right\rceil}_{} = m$$ so that $g$ has a root of multiplicity $2m-1$ at $z=\pm m^2$. Now let $$ f(z)=g(\tfrac{1}{1-z}).$$
Remark: Oops, this as all the desried properties except that $f$ is not continuoius at $1$.