Find the limit of (the square root of 3x - 2, plus the square root of x - 1, minus 3) divided by (x squared minus 4) as x approaches 2.

Step-by-step solution

1. 1

   Substitute $x=2$ to identify the indeterminate form. Numerator: $\sqrt{4}+\sqrt{1}-3=2+1-3=0$. Denominator: $2^2-4=0$. So the limit is of type $\tfrac00$ — direct substitution fails, but the limit may still exist.

2. 2

   Check the hypotheses before differentiating. Both $\sqrt{3x-2}$ and $\sqrt{x-1}$ are differentiable near $x=2$ (their radicands are $4$ and $1$, safely positive), and the denominator's derivative $2x$ is nonzero at $x=2$. l'Hopital's rule therefore applies.

3. 3

   Differentiate numerator and denominator separately.

   $\frac{d}{dx}\left(\sqrt{3x-2}+\sqrt{x-1}-3\right)=\frac{3}{2\sqrt{3x-2}}+\frac{1}{2\sqrt{x-1}},\qquad \frac{d}{dx}\left(x^2-4\right)=2x$

4. 4

   Substitute $x=2$ into the new quotient.

   $\frac{\dfrac{3}{2\sqrt{4}}+\dfrac{1}{2\sqrt{1}}}{2(2)}=\frac{\tfrac34+\tfrac12}{4}=\frac{\tfrac54}{4}=\frac{5}{16}$

5. 5

   Confirm numerically from both sides. Evaluating the original quotient at $x=2.0001$ gives $0.3124855$ and at $x=1.9999$ gives $0.3125145$; both bracket $\tfrac{5}{16}=0.3125$, so the two-sided limit is confirmed.

6. 6

   Note the algebraic alternative. Multiplying by the conjugate of each root and factoring $x^2-4=(x-2)(x+2)$ produces $\dfrac{3}{(x+2)(\sqrt{3x-2}+2)}+\dfrac{1}{(x+2)(\sqrt{x-1}+1)}$, which at $x=2$ is $\tfrac{3}{16}+\tfrac{1}{8}=\tfrac{5}{16}$ — the same value without calculus.