Question:medium

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

Show Hint

To find continuity, equate \( \lim_x \to a f(x) \) and \( f(a) \).
Updated On: Jan 13, 2026
  • \( 0 \)
  • \( \frac14 \)
  • \( 1 \)
  • \( 4 \)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Continuity Condition
For continuity at \( x = 0 \), the limit of \( f(x) \) as \( x \) approaches 0 must equal \( f(0) \), which is \( k \). So, \( \lim_{x \to 0} f(x) = f(0) = k \).

Step 2: Limit Evaluation
For \( x eq 0 \), the function is given by: \[ f(x) = \frac{\sqrt{4} + x - 2x}{x}. \] To evaluate the limit, multiply the numerator and denominator by the conjugate of the numerator's non-radical term, which is \( \sqrt{4} + x + 2 \): \[ f(x) = \frac{(\sqrt{4} + x - 2)(\sqrt{4} + x + 2)}{x(\sqrt{4} + x + 2)} \] This simplifies to: \[ = \frac{(4 + x) - 4}{x(\sqrt{4} + x + 2)} \] \[ = \frac{x}{x(\sqrt{4} + x + 2)} \] For \( x eq 0 \), we can cancel \( x \): \[ = \frac{1}{\sqrt{4} + x + 2}. \]
Step 3: Substitution at \( x = 0 \)
Now, substitute \( x = 0 \) into the simplified expression to find the limit: \[ \lim_{x \to 0} f(x) = \frac{1}{\sqrt{4} + 0 + 2} = \frac{1}{2 + 0 + 2} = \frac{1}{4}. \]
Step 4: Determining \( k \)
From Step 1, we know that \( k = \lim_{x \to 0} f(x) \). Therefore, \( k = \frac{1}{4} \). This matches option (B).
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