We are given the function $f(x) = ax + b$ and the limit equation:

$\lim_{{x \to 4}} \frac{f(x)}{\sqrt{x} - 2} = -4$

We are tasked with finding $f(1)$.

Step 1: Find the value of $a$

We know that $f(x) = ax + b$, so substitute it into the given limit equation:

$\lim_{{x \to 4}} \frac{ax + b}{\sqrt{x} - 2} = -4$

Since $\lim_{{x \to 4}} \sqrt{x} = 2$, we apply L'Hopital's Rule (because the direct substitution leads to an indeterminate form $\frac{0}{0}$):

Differentiate the numerator and denominator:

• The derivative of the numerator $f(x) = ax + b$ is $f'(x) = a$.
• The derivative of the denominator $\sqrt{x} - 2$ is $\frac{1}{2\sqrt{x}}$.

Now, applying the limit again:

$\lim_{{x \to 4}} \frac{a}{\frac{1}{2\sqrt{x}}} = \lim_{{x \to 4}} 2a\sqrt{x} = 4a$

Given that the limit equals $-4$:

$4a = -4 \quad \Rightarrow \quad a = -1$

Step 2: Find the value of $b$

We now use the fact that $f(4) = ax + b$ when $x = 4$:

$f(4) = -1(4) + b = -4 + b$

From the original limit equation, we know that at $x = 4$, the numerator must be 0 for the limit to work, so:

$-4 + b = 0 \quad \Rightarrow \quad b = 4$

Step 3: Calculate $f(1)$

Now that we know $a = -1$ and $b = 4$, substitute into the equation for $f(x)$:

$f(1) = -1(1) + 4 = -1 + 4 = 3$

Thus, the value of $f(1)$ is 3.

The correct answer is (C) 3.

Would you like further details or have any questions? Here are 5 related questions to explore:

1. How do we apply L'Hopital's Rule for different types of limits?
2. What are other common indeterminate forms in calculus?
3. Can we solve this problem using a different limit technique?
4. How does the value of $a$ influence the slope of the line $f(x) = ax + b$?
5. How can we determine limits involving square roots without L'Hopital's Rule?

Tip: L'Hopital's Rule is often useful when limits result in $\frac{0}{0}$ or $\frac{\infty}{\infty}$. Always check for these forms before applying the rule!