In Spivak, Ch. 5 on Limits, problem 3 we are asked to determine the limit $l$ for the given function and value of $a$, and to prove that it is the limit by showing how to find a $\delta$ such that $|f(x)-l|<\epsilon$ for all $x$ satisfying $0<|x-a|<\delta$.
The solution for this problem is not available in the solution manual.
Now consider item $viii$, which specifies the function $f(x)=\sqrt{x}$ near $x=1$
$$|x-1|=|(\sqrt{x})^2-1^2|=|(\sqrt{x}+1)(\sqrt{x}-1)|\leq|\sqrt{x}+1||\sqrt{x}-1|$$
Assume $|\sqrt{x}-1|<1$
$$\iff -1<\sqrt{x}-1<1\implies 1<\sqrt{x}+1<3$$
$$\iff |\sqrt{x}+1|<3\tag{1}$$
$$\iff |\sqrt{x}+1||\sqrt{x}-1|<3|\sqrt{x}-1|$$
$$\iff |x-1|<3|\sqrt{x}-1|<3\epsilon\tag{2}$$
In words: Assume that $f(x)$ is less than 1 unit away from the value $1$. From this follows that $|x-1|<3|\sqrt{x}-1|<3\epsilon$. However, because all the steps involved an $\iff$, we could actually start from assuming $|x-1|<3|\sqrt{x}-1|<3\epsilon$ and reach $|\sqrt{x}-1|<1$.
In any case the conclusion is that
$$\forall \epsilon>0,|x-1|<3\epsilon\implies |\sqrt{x}-1|<\min(1,\epsilon)$$
$$\implies \lim_\limits{x \to 1}\sqrt{x}=1$$
I feel like there is something fishy in these steps from the perspective of adhering strictly to the definition of limit. That being said, it also seems to be correct that every time $f(x)$ gets within $\epsilon$ of $l$, x is within $3\epsilon$ of $1$
In particular, is the use of $\iff$ from $(1)$ to $(2)$ correct?
Note that I think there is an easier solution here, but I am interested in how to make the solution above correct.
The easier solution is
$$|\sqrt{x}-1|=\frac{|\sqrt{x}-1||\sqrt{x}+1|}{|\sqrt{x}+1|}=\frac{|x-1|}{|\sqrt{x}+1|}$$
$$|\sqrt{x}+1|>1 \forall x$$
$$|x-1| < \delta \implies |\sqrt{x}-1|<\delta$$
Therefore
$$\forall \epsilon>0, \exists \delta=\epsilon : |x-1| < \delta \implies |\sqrt{x}-1|<\delta$$
$$\implies \lim_\limits{x\to 1} \sqrt{x}=1$$
Your conclusion $\forall \epsilon>0,|x-1|<3\epsilon\implies |\sqrt{x}-1|<\min (1,\epsilon)$ is incorrect. For example if $\epsilon =1$ and $x=-1.$
Given $e>0,$ we are looking for some $d_e>0$ such that $|x-1|<d_e\implies |\sqrt{x}-1|<e .$ But we also need $x\ge 0,$ which is not guaranteed by $|x-1|<d_e$ unless $d_e\le 1.$
So we begin with the partial condition $d_e\le 1.$
Now if $0<d_e\le 1$ then $|x-1|<d_e\implies 0\le x\implies |\sqrt x -1|=\dfrac {|x-1|}{\sqrt x +1}\le |x-1|$ because the denominator $\sqrt x +1\ge 1$ for any $x\ge 0$. This works if $d_e\le e,$ but we have also included the previous partial condition $d_e\le 1$. So it suffices to put $d_e=\min (1,e).$ Then you have $$\forall e>0\,\forall x\, (|x-1|<d_e=\min (1,e)\implies |\sqrt x -1|<e).$$