Is there any number that I can't find?

Allowing limits is very strong.

A real number $y$ has an integer part and a decimal part $a.x_1x_2x_3x_4...$ We can construct this real number as a limit in the following way:

Let $y_1 = a, y_2=a.x_1, y_3 =a.x_1x_2 , y_4 = a.x_1x_2x_3 , y_5=...$ clearly each $y_i$ is a rational number and $\lim_{n\to\infty} y_i = y$.

Maybe Chaitin's constant is somthing you are looking for.

It is non-computable, so we cannot approximate it to arbitrary precision, even with stronger tools than the ones allowed by you.

Every real number $x$ can be written in the form: $$x=n+\sum_{k=0}^\infty x_k\left(\frac{1}{10}\right)^k,\ x_k\in\{0,1,\dots,9\}$$ and we are used to call $n$ in integer part and $x_k$ its decimal digits. So the answer to you question is yes.

Moreover, you can write any real number $x$ in the form: $$x=n+\sum_{k=1}^\infty b_k\left(\frac{1}{2}\right)^k,\ b_k\in\{0,1\}$$ where $b_k$ are its binary digits.

So, you only need two digits and the operations you've mentioned.