<s>[INST] 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 $:
$$ 2 \cdot \lim _ { x \to 3 } \left( \sqrt x + \sqrt 3 \right) =
2 \left( \sqrt 3 + \sqrt 3 \right) = 4 \sqrt 3 $$
 [/INST] Yes, the solution is correct. The only minor stylistic change I would personally consider is not going into as much detail when factoring out the 2 in the second line, and mention that you're multiplying your fraction by $\frac{\sqrt{x}+\sqrt{3}}{\sqrt{x}+\sqrt{3}}.$
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