`lim_(x to 4) (x^(2) - 16)/(sqrt(x) - 2)` is equal to

Options

A. 8
B. 64
C. 16
D. 32

Answer

To solve the limit \( \lim_{x \to 4} \frac{x^2 - 16}{\sqrt{x} - 2} \), we can follow these steps:

### Step 1: Direct Substitution
First, we can substitute \( x = 4 \) directly into the expression:
\[ \frac{4^2 - 16}{\sqrt{4} - 2} = \frac{16 - 16}{2 - 2} = \frac{0}{0} \]
This gives us an indeterminate form \( \frac{0}{0} \).

**Hint:** When you get \( \frac{0}{0} \), consider using L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule
Since we have an indeterminate form, we can apply L'Hôpital's Rule, which states that:
\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]
if the limit on the right exists.

Here, let \( f(x) = x^2 - 16 \) and \( g(x) = \sqrt{x} - 2 \).

### Step 3: Differentiate the Numerator and Denominator
Now, we differentiate \( f(x) \) and \( g(x) \):
- The derivative of \( f(x) = x^2 - 16 \) is \( f'(x) = 2x \).
- The derivative of \( g(x) = \sqrt{x} - 2 \) is \( g'(x) = \frac{1}{2\sqrt{x}} \).

### Step 4: Rewrite the Limit
Now we can rewrite the limit using the derivatives:
\[ \lim_{x \to 4} \frac{f'(x)}{g'(x)} = \lim_{x \to 4} \frac{2x}{\frac{1}{2\sqrt{x}}} \]

### Step 5: Simplify the Expression
This simplifies to:
\[ \lim_{x \to 4} \frac{2x \cdot 2\sqrt{x}}{1} = \lim_{x \to 4} 4x\sqrt{x} \]

### Step 6: Substitute \( x = 4 \)
Now substitute \( x = 4 \):
\[ 4 \cdot 4 \cdot \sqrt{4} = 4 \cdot 4 \cdot 2 = 32 \]

### Conclusion
Thus, the limit is:
\[ \lim_{x \to 4} \frac{x^2 - 16}{\sqrt{x} - 2} = 32 \]

**Final Answer:** \( 32 \)

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