Solution verification: $ \lim _ { x \to 3 } \frac { 2 x - 6 } { \sqrt x - \sqrt 3 } = ? $ This limit is not too difficult but I was just wondering if my work/solution looked good?
Thanks so much for your input!!
$$ \lim _ { x \to 3 } \frac { 2 x - 6 } { \sqrt x - \sqrt 3 } = ? $$
$$ 2 x - 6 = 2 x \left( 1 - \frac 6 { 2 x } \right) $$
$$ \lim _ { x \to 3 } \frac { 2 x - 6 } { \sqrt x - \sqrt 3 } =
\lim _ { x \to 3 } \frac { 2 x \left( 1 - \frac 6 { 2 x } \right) } { \sqrt x - \sqrt 3 } =
2 \cdot \lim _ { x \to 3 } \frac { x \left( 1 - \frac 6 { 2 x } \right) } { \sqrt x - \sqrt 3 } =
2 \cdot \lim _ { x \to 3 } \frac { x - 3 } { \sqrt x - \sqrt 3 } $$
By rationalizing the denominator:
$$ \frac { x - 3 } { \sqrt x - \sqrt 3 } = \sqrt x + \sqrt 3 $$
$$ 2 \cdot \lim _ { x \to 3 } \frac { x - 3 } { \sqrt x - \sqrt 3 } =
2 \cdot \lim _ { x \to 3 } \left( \sqrt x + \sqrt 3 \right) $$
By plugging in $ x = 3 $:
$$ 2 \cdot \lim _ { x \to 3 } \left( \sqrt x + \sqrt 3 \right) =
2 \left( \sqrt 3 + \sqrt 3 \right) = 4 \sqrt 3 $$

Your solution is ok, but a bit verbose.
To remedy to that, I personally suggest working around zero by setting $x=3+u$ with $u\to 0$, I find it triggers natural reflexes more. Also for presentation purposes, I prefer working on the expression and then make use of $\to$ to specify the limit rather than carrying the $\ \lim\limits_{x\to 3}\ $operator everywhere, and the fact the the limit is now in zero helps a lot (it makes the context obvious).
Compare how much shorter this is:
$\require{cancel}f(x)=\dfrac{2x-6}{\sqrt{x}-\sqrt{3}}=\dfrac{2u}{\sqrt{3+u}-\sqrt{3}}\overset{(*)}{=}\dfrac{2\cancel u}{\cancel u}(\overbrace{\sqrt{3+u}}^{\to\ \sqrt{3}}+\sqrt{3})\to4\sqrt{3}$
$(*)$ multiply by conjugated quantity.
