Question

Let \( f(x) = \frac{2 - \sqrt{x + 4}}{\sin 2x}, \, x \neq 0 \). In order that \( f(x) \) is continuous at \( x = 0 \), \( f(0) \) is to be defined as:

• A. \( -\frac{1}{8} \)
• B. \( \frac{1}{2} \)
• C. \( 1 \)
• D. \( \frac{1}{8} \)

Hint:

Rule is a useful technique for resolving indeterminate limits, particularly when dealing with square roots and trigonometric functions.

Answer 1

The Correct Option is A

For \( f(x) \) to be continuous at \( x = 0 \), it must satisfy \( f(0) = \lim_{x \to 0} f(x) \). Step 1: Evaluate the limit. Given the function \( f(x) = \frac{2 - \sqrt{x + 4}}{\sin 2x} \). As \( x \to 0 \), the expression becomes an indeterminate form \( \frac{0}{0} \). L'Hôpital's Rule is applied: \[\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\frac{d}{dx}(2 - \sqrt{x + 4})}{\frac{d}{dx}(\sin 2x)}.\] Step 2: Differentiate the numerator and denominator. The derivatives are: \[\frac{d}{dx}(2 - \sqrt{x + 4}) = -\frac{1}{2\sqrt{x + 4}}, \quad \frac{d}{dx}(\sin 2x) = 2 \cos 2x.\] Substituting these into the limit: \[\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{-\frac{1}{2\sqrt{x + 4}}}{2 \cos 2x}.\] Step 3: Simplify the expression. At \( x = 0 \): \[\sqrt{x + 4} = \sqrt{4} = 2, \quad \cos 2x = \cos 0 = 1.\] Substitute these values: \[\lim_{x \to 0} f(x) = \frac{-\frac{1}{2 \cdot 2}}{2 \cdot 1} = \frac{-\frac{1}{4}}{2} = -\frac{1}{8}.\] Step 4: Define \( f(0) \). For continuity at \( x = 0 \), \( f(0) \) is defined as: \[f(0) = -\frac{1}{8}.\] Final Answer: \[\boxed{-\frac{1}{8}}.\]