Limit of sequence $a_1=1$ and $a_{n+1}=\frac{2a_n}{4a_n+1}$

Note that if $a_n\geqslant\frac14$, then$$a_{n+1}-\frac14=\frac{2a_n}{4a_n+1}-\frac14=\frac{4a_n-1}{4(4a_n+1)}\geqslant0.$$So, $a_{n+1}\geqslant\frac14$. Since $a_1>\frac14$, this proves that $(\forall n\in\mathbb{N}):a_n\geqslant\frac14$. Therefore, the limit is $\frac14$.

Let $b_n=\frac1{a_n}$ (possible because the recursion formula cannot produce zero from non-zero input). Then $$b_{n+1}=\frac{4a_n+1}{2a_n} =2+\frac1{2a_n}=2+\frac12b_n.$$ I suppose it is easier to see that $b_n\to 4$ (or at least that $b_n$ is bounded and hence $a_n\not\to 0$). Indeed, $$ b_{n+1}-4=-a+\frac12b_n=\frac12(b_n-4)$$ so that $$b_n-4=2^{-n}(b_0-4). $$