Question

find the following limit or state that it does not exist
$$\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4}$$

simplify the given limit

$$\\lim_{x \\to 4} \\frac{\\sqrt{x} - 2}{x - 4} = \\lim_{x \\to 4} \\left( \\frac{1}{\\sqrt{x} + 2} \ ight)$$

(simplify your answer.)

evaluate the limit, if possible. select the correct choice below and, if necessary, fill in the answer box to complete your choice.

○ a. $$\lim_{x \to 4} \frac{\sqrt{x} - 2}{x - 4} = \square$$ (type an exact answer.)

○ b. the limit does not exist.

Explanation:

Step1: Substitute \( x = 4 \) into \( \frac{1}{\sqrt{x}+2} \)

We know that the simplified limit is \( \lim_{x \to 4} \frac{1}{\sqrt{x}+2} \). Now we substitute \( x = 4 \) into the function \( \frac{1}{\sqrt{x}+2} \). First, calculate \( \sqrt{4} \), which is 2. Then the denominator becomes \( 2 + 2 = 4 \). So the function value is \( \frac{1}{4} \).

Step2: Evaluate the limit

Since the function \( \frac{1}{\sqrt{x}+2} \) is continuous at \( x = 4 \) (the denominator is not zero at \( x = 4 \)), the limit as \( x \to 4 \) is equal to the function value at \( x = 4 \), which we calculated as \( \frac{1}{4} \).

Answer:

A. \( \lim\limits_{x \to 4} \frac{\sqrt{x}-2}{x - 4} = \frac{1}{4} \)