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

Step-by-step solution

1. 1

   Confirm the indeterminate form. At $x=2$ the numerator is $\sqrt{6-2}-2 = 2-2 = 0$ and the denominator is $\sqrt2-\sqrt2 = 0$, so it is $\tfrac00$ and direct substitution fails.

2. 2

   Substitute t = root x to make the denominator linear. With $x=t^2$ and $t\to\sqrt2$, $\lim_{t\to\sqrt2}\frac{\sqrt{3t^2-2}-2}{t-\sqrt2}.$ A linear denominator of the form $t-c$ is the signature of a derivative.

3. 3

   Recognise the difference quotient. Put $f(t)=\sqrt{3t^2-2}$. Then $f(\sqrt2)=\sqrt{6-2}=2$, so the expression is exactly $\frac{f(t)-f(\sqrt2)}{t-\sqrt2} \longrightarrow f'(\sqrt2).$

4. 4

   Differentiate with the chain rule. $f'(t) = \frac{1}{2}(3t^2-2)^{-1/2}\cdot 6t = \frac{3t}{\sqrt{3t^2-2}}.$

5. 5

   Evaluate the derivative at t = root 2. $f'(\sqrt2) = \frac{3\sqrt2}{\sqrt{3\cdot2-2}} = \frac{3\sqrt2}{2} \approx 2.121320.$

6. 6

   Cross-check by rationalising instead. Multiplying by $\dfrac{\sqrt{3x-2}+2}{\sqrt{3x-2}+2}\cdot\dfrac{\sqrt x+\sqrt2}{\sqrt x+\sqrt2}$ turns the quotient into $\dfrac{3(x-2)}{x-2}\cdot\dfrac{\sqrt x+\sqrt2}{\sqrt{3x-2}+2}$, which at $x=2$ gives $3\cdot\dfrac{2\sqrt2}{4} = \dfrac{3\sqrt2}{2}$ - the same value. Evaluating the original at $x=2.000001$ gives $2.1213202$, confirming it numerically.