Given the limit expression:

$\lim_{x \to 2} \frac{a\sqrt{x} + 7 + b}{x - 2} = 3$

We need to determine the value of $3a + b$.

Step 1: Substitution of the limit

Since we are evaluating the limit as $x \to 2$, directly substituting $x = 2$ into the numerator and denominator yields:

$\frac{a\sqrt{2} + 7 + b}{2 - 2} = \frac{a\sqrt{2} + 7 + b}{0}$

This gives a division by zero, which implies that the limit is undefined unless the numerator also tends to zero as $x \to 2$. Therefore, we require:

$a\sqrt{2} + 7 + b = 0$

Thus, the equation is:

$a\sqrt{2} + b = -7 \quad \text{(Equation 1)}$

Step 2: Apply L'Hopital's Rule

Since we have a $\frac{0}{0}$ form, we can apply L'Hopital's Rule by differentiating the numerator and denominator with respect to $x$.

• Derivative of the numerator $a\sqrt{x} + 7 + b$ is:

$\frac{d}{dx}(a\sqrt{x} + 7 + b) = a \cdot \frac{1}{2\sqrt{x}}$

• Derivative of the denominator $x - 2$ is:

$\frac{d}{dx}(x - 2) = 1$

Thus, after applying L'Hopital's Rule, the limit becomes:

$\lim_{x \to 2} \frac{a \cdot \frac{1}{2\sqrt{x}}}{1} = 3$

Substitute $x = 2$:

$\frac{a \cdot \frac{1}{2\sqrt{2}}}{1} = 3$

This simplifies to:

$\frac{a}{2\sqrt{2}} = 3$

Solve for $a$:

$a = 6\sqrt{2} \quad \text{(Equation 2)}$

Step 3: Substitute $a$ into Equation 1

Substitute $a = 6\sqrt{2}$ into $a\sqrt{2} + b = -7$:

$6\sqrt{2} \cdot \sqrt{2} + b = -7$

This simplifies to:

$12 + b = -7$

Solve for $b$:

$b = -19$

Step 4: Calculate $3a + b$

Now, we calculate $3a + b$:

$3a + b = 3(6\sqrt{2}) + (-19) = 18\sqrt{2} - 19$

Thus, the value of $3a + b$ is $18\sqrt{2} - 19$.

---

Would you like any further details or clarifications on this process?

Here are 5 related questions to explore further:

1. How does L'Hopital's Rule help in evaluating limits?
2. What are the conditions for applying L'Hopital's Rule?
3. How do you interpret limits involving square roots?
4. What are common indeterminate forms, and how are they resolved?
5. How can we verify if a limit tends to infinity or a finite number?

Tip: Always check if you have a $\frac{0}{0}$ form before applying L'Hopital's Rule to avoid mistakes.