x  c(x) 
2  0.693147... 
?  1.000000... 
3  1.098613... 
4  1.386295... 
5  1.609439... 

There is a reason why taking the square root halves the
rubbish  it's because $(1+\epsilon/2)^2$ is $1+\epsilon+\epsilon^2/4$,
and when $\epsilon$ is small enough, $\epsilon^2$ effectively vanishes.
So $(1+\epsilon/2)^2\approx 1+\epsilon$, provided $\epsilon$ is small.
Hence $\sqrt{1+\epsilon} \approx 1+\epsilon/2$


Take 10, and square root many times over. After a while you get 1 +
rubbish. Taking the square root of that halves the rubbish, so we
can deduce that $10^\epsilon = 1+c_{10}\epsilon$.
I've put the subscript on the $c$ because if you do this again but
start with 5 instead of 10 you get a different constant, $c_5$.
So now we have a function, $c(x)$ where you start with $x$,
successively square root, compute the constant, and that's
the value of the function.
If we do this there are lots of questions to ask about the
function $c$:
 Is it well defined?
 Is it continuous?
and so on. But we can observe that $c(2)<1$ and $1<c(3)$, and
probably $c(2)<c(x)<c(3)$ when $x$ is between 2 and 3, so is
there a value of $x$ that gives $c(x)=1$?
The answer is yes, the answer is $e$.
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