Solution verification: $ \lim _ { x \to 3 } \frac { 2 x - 6 } { \sqrt x - \sqrt 3 } = ? $

Question

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 $:

$$ 2 \cdot \lim _ { x \to 3 } \left( \sqrt x + \sqrt 3 \right) = 2 \left( \sqrt 3 + \sqrt 3 \right) = 4 \sqrt 3 $$

Answer 1

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 $\frac{\sqrt{x}+\sqrt{3}}{\sqrt{x}+\sqrt{3}}.$

Answer 2

Shortly: for $x\ne\sqrt 3$,

$$\frac{2x-6}{\sqrt x-\sqrt3}=2\frac{(\sqrt x-\sqrt3)(\sqrt x+\sqrt3)}{\sqrt x-\sqrt3}=2(\sqrt x+\sqrt3)\to4\sqrt3.$$

Answer 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.