Given a continuous real function $f:[0,1]\to\mathbb{R}$, for $k\in\mathbb{N}$, we need to find a rational polynomial $p$ satisfying $p^{(i)}(0)=0$ $(1\leq i\leq k-1)$ such that for $\epsilon>0$, $\Vert f-p\Vert < \epsilon$. ($p^{(i)}(x)$ is the derivative of order $i$.)
Using the Stone-Weierstrass theorem, I proved that for we can find a rational polynomial that approximates a continuous function $f$. Clearly, the given polynomial is of the form $p(x)= a_0 +x^kq(x)$, where $q$ is another rational polynomial.
I defined $$g(x) := \frac{f(x)-f(0)}{x^k}$$ for which we can find an approximate rational polynomial, and hence we can find a rational polynomial that approximates $f$ satisfying the above property. However, $g$ is not necessarily continuous at $0$, and hence we cannot simply use Weierstrass approximation. Any hints?
For $k\in \mathbb N,$ let $A_k$ denote the polynomials on $[0,1]$ that have the form $a+x^kp(x),$ where $p$ is a polynomial. Then each $q\in A_k$ satisfies $q^{(j)}(0)=0,$ $j=1,\dots, k-1.$
Furthermore each $A_k$ is an algebra (easy exercise) that vanishes nowhere (because $1\in A_k$) and separates points (because $x^k\in A_k$). By Stone-Weierstrass, $A_k$ is dense in $C([0,1]).$
So to reach the desired conclusion, all we need to show is that each $q\in A_k$ is the uniform limit of polynomials of the form $a+x^kp(x),$ where $a\in \mathbb Q$ and $p$ is a rational polynomial. I'll leave that to you for now.