Epsilon-delta proofs $\newcommand{\absval}[1]{\left\lvert #1 \right\rvert}$
I am self-learning Real Analysis from Understanding Analysis by Stephen Abott. I am getting back to $\delta-\epsilon$ arguments after a break, so I'd like ask if my $\delta$-response to the below $\epsilon$-challenges is technically correct and rigorous.

Problem 4.2.2. For each stated limit, find the largest possible $\delta-$neighbourhood that is a proper response to the given $\epsilon$-challenge.


(a) $\lim_{x \to 3}(5x - 6) = 9$, where $\epsilon = 1$


(b) $\lim_{x \to 4}\sqrt{x} = 2$, where $\epsilon = 1$


(c) $\lim_{x \to \pi} [[x]] = 3$, where $\epsilon = 1$. (The function $[[x]]$ returns the greatest integer less than or equal to $x$)


(d) $\lim_{x \to \pi} [[x]] = 3$, where $\epsilon = 0.01$

Proof.
(a) We would like to make the distance $\absval{(5x - 6) - 9}<1$, so $\absval{5x - 15} < 1$, thus $\absval{x - 3} < 1/5$. Thus, $\delta = 1/5$.
(b) The expression $\absval{\sqrt{x} - 2}$ can be written as,
\begin{align*}
 \sqrt{x} - 2 &= (\sqrt{x} - 2) \times \frac{(\sqrt{x} + 2)}{(\sqrt{x} + 2)}\\
 &= \frac{\absval{(x - 4)}}{\absval{\sqrt{x} + 2}}\\
 &\le \frac{\absval{(x - 4)}}{2}\\
 &< \frac{\delta}{2}
\end{align*}
So, if we pick $\delta = 2$, the distance $\absval{\sqrt{x} - 2}$ would be smaller than $\epsilon = 1$.
(c) Consider the expression $\absval{[[x]] - 3}$.
We are interested to make the distance $\absval{[[x]] - 3}$ smaller than $\epsilon = 1$. So, $2 < [[x]] < 4$. Therefore, $[[x]]=3$. This is true, if and only if, $3 \le x < 4$. Consequently, $3 - \pi \le x - \pi < 4 - \pi$. This inequality is satisfied, if the distance $\absval{x - \pi} < \pi - 3$. So, $\delta = \pi - 3$.
(d) Again, $\delta = \pi - 3$ is a proper response to the given $\epsilon-$ challenge.

all good except part (b), which is wrong because you are requested to find the largest $\delta$, which must be $(\delta = 3)$, since
$|\sqrt{4-3} - 2| = 1.$
