There is also the useful notion of left-computability and right-computability.
A real number $x$ is left-computable if for any computable real $a$ you can answer YES within finite time to the question : is $x$ greater than $a$ ?
A real number $x$ is right-computable if $-x$ is left-computable.
A real number $x$ is computable if and only if it is left and right-computable.
A real number $x$ is left-computable if and only if an algorithm can give you a non-decreasing sequence of rational numbers which tends to $x$.
For example, given a polynomial with real computable coefficients. Then its largest root is not a computable function of the coefficients. However, it is right-computable.
Following the friendly controversy with Carl Mummert, I realized that I was unclear.
As an example, consider the polynomial $f(x)=(x+1)x^2-\epsilon$, with $\epsilon$ a small computable real number. The largest root of $f$ is near $\sqrt\epsilon$ if $\epsilon\geq 0$, and near $-1$ if not.
Given the information that $\epsilon=0$ or not, this root is computable. However, as a function of $\epsilon$, it is not. If it were, it would imply that we are able to decide if $\epsilon\geq 0$ or not. But we can't : if you make run the Turing machine which approximate $\epsilon$ and it gives you only zero, you will never be able to state after a finite number of steps that $\epsilon$ is positive, or negative, or null.
This problem is quite general, and in fact, the Grzegorczyk states that a computable function from real numbers to real numbers is continuous. And obviously the function of $\epsilon$ "largest root of f" is not.
However, this largest root is right-computable : the algorithm answer zero as long as it as not enough precision to disprove $\epsilon=0$, and then it is able chose the correct location. But if the algorithm answers zero, you can never be sure that there is indeed a root near zero.
Carl Mummert, please correct me if I'm wrong.