Question

If f(x)= \(\frac{\sqrt{4} + x - 2}{x}, If \ x \neq 0 \\ k \ If \ x \neq 0\) ,is continuous at x = 0, then the value of k is:

• A. 0
• B. 4
• C. \(\frac{1}{4}\)
• D. 1

Answer 1

The Correct Option is C

To determine the value of \(k\) for which the function \(f(x)\) is continuous at \(x=0\), we need to ensure that \(\lim_{{x \to 0}} f(x) = f(0)\). Since \(f(x)\) is defined in two parts, for \(x \neq 0\), \(f(x) = \frac{\sqrt{4} + x - 2}{x}\) and for \(x=0\), \(f(x) = k\).

First, simplify \(\frac{\sqrt{4} + x - 2}{x}\) for \(x \neq 0\):

Given: \(\sqrt{4} = 2\), thus the function becomes \(\frac{2 + x - 2}{x} = \frac{x}{x} = 1\) for \(x \neq 0\).

Now, compute the limit as \(x\) approaches 0:

\(\lim_{{x \to 0}} f(x) = \lim_{{x \to 0}} 1 = 1\).

Since \(f(x)\) must be continuous at \(x=0\), \(\lim_{{x \to 0}} f(x) = f(0)\), leading to:

\(1 = k\).

Thus, the function \(f(x)\) is continuous at \(x=0\) if \(k = 1\).