Prove $f(x) = \sqrt{x}$ is continuous at $x = 4$ Show that $f(x) = \sqrt{x}$ is continuous at $x = 4$
So my textbook has a proof for this and this is their scratch work:
$\forall\epsilon>0$ $\exists\delta>0$ s.t $\forall x$ $0<|x-4|<\delta \implies |\sqrt{x}-2|<\epsilon$
Working backwards:
$$|\sqrt{x}-2|<\epsilon \iff \frac{1}{\sqrt{x}+2}|x-4| < \epsilon$$
then they stated that 'let us assume that $\delta\le4$'
$$|x-4|<\delta \implies |x-4|<4 ...$$
My question is why did they assume that $\delta\le4$ (there was no explaintion in the text) instead of the normal cases when you assume that $\delta\le1$?
 A: The goal of choosing an upper bound is to deal with the $\frac{1}{\sqrt{x} + 2}$ term; we want to ensure that this term doesn't get too big.
Since it can't possibly be bigger than $\frac{1}{2}$, all we really need to worry about is to ensure that $x > 0$, so that the function is actually defined.
Insisting that $\delta \leq 4$ works for this purpose, but we could have chosen any smaller value as well. (it's always okay to make $\delta$ smaller than it 'needs' to be) 
For other dealing with sorts of terms, usually any choice of $\delta$ is fine. Sometimes there are exceptions: e.g. if the extra term to deal with was $\frac{1}{\sqrt{x} - 1}$, then insisting on $x > 1$ isn't good enough, because this term can become arbitrarily large. However, insisting that $x > 1.1$ is fine (which can be achieved by insisting $\delta \leq 2.9$).
A: If $\delta \gt 4$ then there are negative $x$ which satisfy $|x-4|\lt \delta$ but for which there is not real $\sqrt{x}$.
Using "let us assume $\delta \le 1$" would in fact have worked here, but would not have worked if the original question had been "Prove $f(x)=\sqrt{x}$ is continuous at $x=\frac12$."   So it is reasonable to assume an upper bound for $\delta$ which relates to the actual question. 
