Question

If \[ f(9) = 0 \quad \text{and} \quad f'(9) = 0, \] then \[ \lim_{x \to 9} \frac{\sqrt{f(x)} - 3}{\sqrt{x} - 3} \] is equal to: 

• A. \( 0 \)
• B. \( f(0) \)
• C. \( f'(3) \)
• D. \( f(9) \)
• E. \( 1 \)

Hint:

L'Hôpital's Rule is the most efficient way to handle limits involving square roots and unknown functions, provided the functions are differentiable at the limit point.

Answer 1

The Correct Option is A

Concept: Direct substitution into the limit yields a potential indeterminate form. Given the conditions \( f(9)=0 \) and \( f'(9)=0 \), we apply L'Hôpital's Rule, which allows us to differentiate the numerator and denominator separately when a limit results in \( 0/0 \) or \( \infty/\infty \).

Step 1: Differentiating using the Chain Rule.
Applying L'Hôpital's Rule to the expression: \[ L = \lim_{x \to 9} \frac{\frac{d}{dx}(\sqrt{f(x)} - 3)}{\frac{d}{dx}(\sqrt{x} - 3)} \] \[ L = \lim_{x \to 9} \frac{\frac{1}{2\sqrt{f(x)}} \cdot f'(x)}{\frac{1}{2\sqrt{x}}} = \lim_{x \to 9} \frac{f'(x) \cdot \sqrt{x}}{\sqrt{f(x)}} \]

Step 2: Substituting the given values.
We substitute \( x = 9 \), \( f(9) = 0 \), and \( f'(9) = 0 \). Because the derivative \( f'(9) \) is zero and the denominator \( \sqrt{f(9)} \) is also zero, we evaluate the limit's behavior. In this context, the numerator's zero dominates the expression: \[ L = \frac{0}{\sqrt{0}} \cdot 3 \rightarrow 0 \]