If `f(x){{:((sqrt(4+x)-2)/(x)",",xne0),(k,","x=0):}` is continuous at x =0 , then the value of k is

Options

A. 1
B. 4
C. `(1)/(4)`
D. 0

Answer

To find the value of \( k \) such that the function

\[ f(x) = \begin{cases} \frac{\sqrt{4+x} - 2}{x} & \text{if } x \neq 0 \\ k & \text{if } x = 0 \end{cases} \]

is continuous at \( x = 0 \), we need to ensure that

\[ \lim_{x \to 0} f(x) = f(0) = k. \]

### Step 1: Calculate the limit as \( x \) approaches 0

We start by evaluating the limit:

\[ \lim_{x \to 0} \frac{\sqrt{4+x} - 2}{x}. \]

### Step 2: Substitute \( x = 0 \)

Substituting \( x = 0 \) directly into the expression gives us:

\[ \frac{\sqrt{4+0} - 2}{0} = \frac{2 - 2}{0} = \frac{0}{0}, \]

which is an indeterminate form. Therefore, we can apply L'Hôpital's Rule.

### Step 3: Apply L'Hôpital's Rule

According to L'Hôpital's Rule, we differentiate the numerator and the denominator:

- The derivative of the numerator \( \sqrt{4+x} - 2 \) is:

\[ \frac{d}{dx}(\sqrt{4+x}) = \frac{1}{2\sqrt{4+x}}. \]

- The derivative of the denominator \( x \) is:

\[ \frac{d}{dx}(x) = 1. \]

Now we can rewrite the limit:

\[ \lim_{x \to 0} \frac{\sqrt{4+x} - 2}{x} = \lim_{x \to 0} \frac{\frac{1}{2\sqrt{4+x}}}{1} = \lim_{x \to 0} \frac{1}{2\sqrt{4+x}}. \]

### Step 4: Evaluate the limit

Now we substitute \( x = 0 \):

\[ \lim_{x \to 0} \frac{1}{2\sqrt{4+x}} = \frac{1}{2\sqrt{4}} = \frac{1}{2 \cdot 2} = \frac{1}{4}. \]

### Step 5: Set the limit equal to \( k \)

Since we want the function to be continuous at \( x = 0 \), we set:

\[ k = \lim_{x \to 0} f(x) = \frac{1}{4}. \]

### Conclusion

Thus, the value of \( k \) is

\[ \boxed{\frac{1}{4}}. \]

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