Possible Error in Elementary Analysis by Ross I am having some serious trouble understanding example 5 from chapter 8 of Elementary Analysis by Ross. The example is to do with proving sequence are convergent, and here is the definition of convergence that the example uses:

A sequence ($s_n$) is said to converge to the real number $s$ provided that for each $\epsilon>0$ there exists a number $N$ such that $n>N$ implies $|s_n-s|<\epsilon$.

Now, the example is as follows:

Let ($s_n$) be a sequence of nonnegative real numbers and suppose $s=\text{lim}s_n$. Note that $s \geq 0$. Prove that lim$\sqrt{s_n}=$lim$\sqrt{s}$.

Next, the author divides the work into two cases: when $s=0$ and when $s>0$. The case in which $s=0$ is left as an exercise and just by inspection appears trivial, but my question is regarding when $s>0$.
He shows that for $\epsilon>0$ we must prove $\exists N \in \mathbb{N}$ such that $n>N$ implies $|\sqrt{s_n}-\sqrt{s}|<\epsilon$
For $s>0$, we have that
$$|\sqrt{s_n}-\sqrt{s}|=\frac{|\sqrt{s_n}-\sqrt{s}||\sqrt{s_n}+\sqrt{s}|}{|\sqrt{s_n}+\sqrt{s}|}=\frac{|s_n-s|}{\sqrt{s_n}+\sqrt{s}}\leq\frac{|s_n-s|}{\sqrt{s}},$$
So we will select $N$ so that $|s_n-s|<\sqrt{s}\epsilon$ for $n>N$.
What I am confused about is as to where he got $|s_n-s|<\sqrt{s}\epsilon$. If you reorganize the equation above, you get $\sqrt{s}|\sqrt{s_n}-\sqrt{s}|\leq|s_n-s|$. I don't understand how $|s_n-s|<\sqrt{s}\epsilon$ when $\sqrt{s}|\sqrt{s_n}-\sqrt{s}|\leq|s_n-s|$ and $|s_n-s|<\epsilon$.
Thanks for any clarification. There is a good chance I misunderstood something up to this point so any help is much appreciated.
 A: What I am confused about is as to where he got $|_−|<\sqrt{s} \epsilon$.

First, what he should have said was that for any $\epsilon' > 0$, we can define $\epsilon = \sqrt{s}\epsilon'$. Now the assumption in your first highlighted box shows that there's some number $N$ such that $n >  N$ implies 
$$
|s_n - s | < \epsilon = \sqrt{s} \epsilon'
$$
Now since $\epsilon'$ was an arbitrary variable name, we can replace it with the name $\epsilon$, and conclude that for any $\epsilon$, there's a number $N$ such that $n > N$ implies 
$$
|s_n - s | < \sqrt{s}\epsilon.
$$
Why did the author choose to establish this apparently odd fact? Because it's the fact that'll make it possible to derive the conclusion the author wants, namely that for any $\epsilon$, there's an $N$ such that $n > N$ implies
$$
|\sqrt{s_n} - \sqrt{s} | < \epsilon.
$$
How'd the author know to do this? The author probably worked backwards to find it; it's how a lot of proofs like this are developed. 
A: The important thing to realise here (and in many other problems from all areas of mathematics) is that there is a huge difference between finding a proof and logically writing a proof.  It is common (though not obligatory) to find a proof by "working backwards": start with what you want to prove, and find the conditions which make it true.  Logically, this is nonsense: you can't prove something is true if you begin by assuming that it is true.  So you then have to start with the conditions you have found, and write a proof which leads to the conclusion you want.
In the present case, the author has, in effect, solved the desired inequality
$$\frac{|s-s_n|}{\sqrt s}<\epsilon$$
to get
$$|s-s_n|<\sqrt s\epsilon\ ;$$
and has noted that since $\sqrt s\epsilon$ is a positive number$^{\textstyle*}$ and $s_n\to s$, there exists $N$ such that this is true whenever $n>N$.  What really should be done now (but he has left it up to the reader), is to write the proof in full and in a clear, correct logical sequence, perhaps as follows.

Let $\epsilon>0$.
Then $\sqrt s\epsilon>0$; therefore there exists $N$ such that if $n>N$, then $|s_n-s|<\sqrt s\epsilon$.
For such $N$, if $n>N$ we have
  $$|\sqrt{s_n}-\sqrt s|\cdots\langle\hbox{algebra as above}\rangle\cdots<\epsilon\ ,$$
  and so by definition $\lim_{n\to\infty}\sqrt{s_n}=\sqrt s$.

So to sum up: the author did not (really) say
$$|\sqrt{s_n}-\sqrt{s}|\leq\frac{|s_n-s|}{\sqrt{s}}\ \hbox{and}\ |s_n-s|<\epsilon
  \quad\Rightarrow\quad |s_n-s|<\sqrt{s}\epsilon\ ,$$
he really said
$$|\sqrt{s_n}-\sqrt{s}|\leq\frac{|s_n-s|}{\sqrt{s}}\ \hbox{and}\ |s_n-s|<\sqrt{s}\epsilon
  \quad\Rightarrow\quad |\sqrt{s_n}-\sqrt{s}|\le\epsilon\ .$$
This way of writing things is not really accurate (or to be kinder, it is accurate but not yet complete) but unfortunately is quite common.

$^{\textstyle*}$ Incidentally, this is why the case $s=0$ has to be treated separately.
A: Let $\epsilon>0$. Take the number $\epsilon\sqrt{s}$. Note that $\epsilon>0$ and $\sqrt{s}>0$ because $s>0$. Then $\epsilon\sqrt{s}>0$. For this positive number, by the fact that $s_n$ converges to $s$, by definition, there exist $N\in\mathbb{N}$ such that if $n>N$ then $|s_n-s|<\epsilon\sqrt{s}$ (if a sequence converges to a number, then, we can make the distance between the sequence and the limit smaller than any positive number and here, $\epsilon\sqrt{s}$ is a positive number). Then
$$|\sqrt{s_n}-\sqrt{s}|=\frac{|\sqrt{s_n}-\sqrt{s}||\sqrt{s_n}+\sqrt{s}|}{|\sqrt{s_n}+\sqrt{s}|}=\frac{|s_n-s|}{\sqrt{s_n}+\sqrt{s}}\leq\frac{|s_n-s|}{\sqrt{s}}<\frac{\epsilon\sqrt{s}}{\sqrt{s}}=\epsilon$$Thus, if $n>N$ then $|\sqrt{s_n}-\sqrt{s}|<\epsilon$.
A: The author did not get $\lvert s_n-s\rvert<\sqrt s\,\varepsilon$. Instead, he used the fact the he knew that, for any $\varepsilon'>0$, there is a natural $N$ such that $n\geqslant N\implies\lvert s_n-s\rvert<\varepsilon'$ and he decided to take $\varepsilon'=\sqrt s\,\varepsilon$. And he did this because that was useful for the proof.
