Evaluate the limit of (sqrt(3x - 2) + sqrt(x - 1) - 3) / (x^2 - 4) as x approaches 2.

Step-by-step solution

1. 1

   Substitute $x=2$ first to identify the indeterminate form. Always check before applying any technique:

   $\text{numerator}=\sqrt{3(2)-2}+\sqrt{2-1}-3=\sqrt4+\sqrt1-3=2+1-3=0$

   $\text{denominator}=2^2-4=0$

   So the expression is of the form $\tfrac00$. This is good news, not bad: it guarantees that the factor $(x-2)$ hides in both the top and the bottom and can be cancelled. Note also that $x=2$ sits comfortably inside the domain, since $3x-2>0$ and $x-1>0$ near $x=2$.

2. 2

   Split the numerator so each square root loses its own value. The constant $-3$ is exactly $-(2+1)$, i.e. the sum of the two radicals evaluated at $x=2$. Rewriting it that way lets each radical be handled separately:

   $\sqrt{3x-2}+\sqrt{x-1}-3=\underbrace{\big(\sqrt{3x-2}-2\big)}_{\to\,0}+\underbrace{\big(\sqrt{x-1}-1\big)}_{\to\,0}$

   This is the key move. Trying to rationalise the three-term numerator in one go creates a mess; splitting it turns one hard limit into two easy conjugate problems.

3. 3

   Rationalise each piece with its conjugate. Multiplying by $\frac{\sqrt{\;}+\text{value}}{\sqrt{\;}+\text{value}}$ turns the difference of square roots into a difference of squares:

   $\sqrt{3x-2}-2=\frac{(3x-2)-4}{\sqrt{3x-2}+2}=\frac{3(x-2)}{\sqrt{3x-2}+2}$

   $\sqrt{x-1}-1=\frac{(x-1)-1}{\sqrt{x-1}+1}=\frac{x-2}{\sqrt{x-1}+1}$

   Both pieces now carry an explicit factor of $(x-2)$ — the very factor that also lives in the denominator.

4. 4

   Cancel the common factor $(x-2)$. Factor the denominator as a difference of squares, $x^2-4=(x-2)(x+2)$, and combine:

   $\frac{\sqrt{3x-2}+\sqrt{x-1}-3}{x^2-4}=\frac{(x-2)\left[\dfrac{3}{\sqrt{3x-2}+2}+\dfrac{1}{\sqrt{x-1}+1}\right]}{(x-2)(x+2)}$

   $=\frac{1}{x+2}\left[\frac{3}{\sqrt{3x-2}+2}+\frac{1}{\sqrt{x-1}+1}\right]$

   Cancelling is legal because the limit only cares about $x\ne2$.

5. 5

   Substitute $x=2$ into the cancelled expression. Nothing is indeterminate any more:

   $\frac{1}{4}\left[\frac{3}{2+2}+\frac{1}{1+1}\right]=\frac14\left[\frac34+\frac12\right]=\frac14\cdot\frac54=\frac{5}{16}$

6. 6

   Cross-check with L'Hopital's rule. Because the form was $\tfrac00$, differentiating top and bottom separately is also valid:

   $\lim_{x\to2}\frac{\dfrac{3}{2\sqrt{3x-2}}+\dfrac{1}{2\sqrt{x-1}}}{2x}=\frac{\tfrac{3}{4}+\tfrac12}{4}=\frac{\tfrac54}{4}=\frac{5}{16}$

   Both routes agree. A numerical spot check confirms it too: at $x=2.00001$ the quotient is $0.3124986\ldots$, and $\tfrac{5}{16}=0.3125$.