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 $\delta\le1$?

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. 
