Consider the function p(1-p) for 0<=p<=1. Using calculus you can see that at p=1/2 it is 1/4 which is the maximum value. If you can see that this is for the binomial related to the standard deviation of the estimate of the proportion which is sqrt(p(1-p)/n) then p=1/2 is the maximum. When p=1 or 0 the standard error is 0 because you will always get all 1s or all 0s respectively. So as you get close to 0 or 1 a continuity argument says that the standard error approaches 0 as p approaches 0 or 1. In fact it decreases monotonically as p approaches 0 or 1. For large n the estimated proportion should be close to the actual proportion.

Consider the function $p(1-p)$ for $0\leq p\leq 1$. Using calculus you can see that at $p=1/2$ the function is equal to $1/4$, which is the maximum value. If you can see that this is for the binomial related to the standard deviation of the estimate of the proportion which is $\sqrt{p(1-p)/n}$ then $p=1/2$ is the maximum. When $p=1$ or $p=0$ the standard error is 0 because you will always get all 1s or all 0s respectively. So as you get close to 0 or 1 a continuity argument says that the standard error approaches 0 as $p$ approaches 0 or 1. In fact, it decreases monotonically as $p$ approaches 0 or 1. For large $n$ the estimated proportion should be close to the actual proportion.

Consider the function p(1-p) for 0<=p<=1. Using calculus you can see that at p=1/2 it is 1/4 which is the maximum value. If you can see that this is for the binomial related to the standard deviation of the estimate of the proportion which is sqrt(p(1-p)/n) then p=1/2 is the maximum. When p=1 or 0 the standard error is 0 because you will always get all 1s or all 0s respectively. So as you get close to 0 or 1 a continuity argument says that the standard error approaches 0 as p approaches 0 or 1. In fact it decreases monotonically as p approaches 0 or 1.

Consider the function p(1-p) for 0<=p<=1. Using calculus you can see that at p=1/2 it is 1/4 which is the maximum value. If you can see that this is for the binomial related to the standard deviation of the estimate of the proportion which is sqrt(p(1-p)/n) then p=1/2 is the maximum. When p=1 or 0 the standard error is 0 because you will always get all 1s or all 0s respectively. So as you get close to 0 or 1 a continuity argument says that the standard error approaches 0 as p approaches 0 or 1. In fact it decreases monotonically as p approaches 0 or 1. For large n the estimated proportion should be close to the actual proportion.