Question

For what value of \( k \), the function given below is continuous at \( x = 0 \)?
\[ f(x) = \begin{cases} \frac{\sqrt{4} + x - 2x}{x}, & x \neq 0 \\ k, & x = 0 \end{cases} \]

• A. \( 0 \)
• B. \( \frac14 \)
• C. \( 1 \)
• D. \( 4 \)

Hint:

To find continuity, equate \( \lim_x \to a f(x) \) and \( f(a) \).

Answer 1

The Correct Option is B

Step 1: Condition for continuity
For continuity at \( x = 0 \), \( \lim_{x \to 0} f(x) = f(0) = k \).

Step 2: Evaluate the limit
For \( x \neq 0 \): \[ f(x) = \frac{\sqrt{4} + x - 2x}{x}. \] Multiply numerator and denominator by \( \sqrt{4} + x + 2 \): \[ f(x) = \frac{(\sqrt{4} + x - 2)(\sqrt{4} + x + 2)}{x(\sqrt{4} + x + 2)} \] \[ = \frac{4 + x - 4}{x(\sqrt{4} + x + 2)} \] \[ = \frac{1}{\sqrt{4} + x + 2}. \]
Step 3: Substitute \( x = 0 \)
\[ \lim_{x \to 0} f(x) = \frac{1}{\sqrt{4} + 0 + 2} = \frac{1}{4}. \]
Step 4: Set \( k = \lim_{x \to 0} f(x) \)
Thus, \( k = \frac{1}{4} \), which matches option (B).