`f(x) = {{:((1- cos 4x)/x^(2), x lt 0),(=a, x =0),(=sqrt(x)/(sqrt(16+sqrt(x))-4), x gt 0):}` If the function be continuous at x = 0, then a =

Options

A. 4
B. 6
C. 8
D. 10

Answer

To determine the value of \( a \) such that the function \( f(x) \) is continuous at \( x = 0 \), we need to ensure that the left-hand limit and right-hand limit at \( x = 0 \) are equal to \( f(0) \).

The function is defined as follows:

\[ f(x) = \begin{cases} \frac{1 - \cos(4x)}{x^2} & \text{if } x < 0 \\ a & \text{if } x = 0 \\ \frac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4} & \text{if } x > 0 \end{cases} \]

### Step 1: Calculate the left-hand limit as \( x \) approaches 0 from the left (\( x \to 0^- \))

We need to find:

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{1 - \cos(4x)}{x^2} \]

Using the trigonometric identity \( 1 - \cos(x) = 2 \sin^2\left(\frac{x}{2}\right) \), we can rewrite:

\[ 1 - \cos(4x) = 2 \sin^2(2x) \]

Thus, we have:

\[ \lim_{x \to 0^-} \frac{1 - \cos(4x)}{x^2} = \lim_{x \to 0^-} \frac{2 \sin^2(2x)}{x^2} \]

### Step 2: Rewrite the limit

We can express \( \sin^2(2x) \) in terms of \( x \):

\[ \lim_{x \to 0^-} \frac{2 \sin^2(2x)}{x^2} = \lim_{x \to 0^-} \frac{2 \sin^2(2x)}{(2x)^2} \cdot 4 = 4 \lim_{x \to 0^-} \frac{\sin^2(2x)}{(2x)^2} \]

### Step 3: Apply the limit property

Using the limit property \( \lim_{u \to 0} \frac{\sin(u)}{u} = 1 \):

\[ \lim_{x \to 0^-} \frac{\sin^2(2x)}{(2x)^2} = 1 \]

Thus, we have:

\[ \lim_{x \to 0^-} \frac{1 - \cos(4x)}{x^2} = 4 \cdot 1 = 4 \]

### Step 4: Calculate the right-hand limit as \( x \) approaches 0 from the right (\( x \to 0^+ \))

Now we calculate:

\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4} \]

To simplify this limit, we can multiply the numerator and denominator by the conjugate of the denominator:

\[ \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{16 + \sqrt{x}} + 4)}{(\sqrt{16 + \sqrt{x}} - 4)(\sqrt{16 + \sqrt{x}} + 4)} = \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{16 + \sqrt{x}} + 4)}{(\sqrt{16 + \sqrt{x}})^2 - 4^2} \]

This simplifies to:

\[ \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{16 + \sqrt{x}} + 4)}{16 + \sqrt{x} - 16} = \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{16 + \sqrt{x}} + 4)}{\sqrt{x}} = \lim_{x \to 0^+} (\sqrt{16 + \sqrt{x}} + 4) \]

As \( x \to 0 \), \( \sqrt{16 + \sqrt{x}} \to \sqrt{16} = 4 \):

\[ \lim_{x \to 0^+} (\sqrt{16 + \sqrt{x}} + 4) = 4 + 4 = 8 \]

### Step 5: Set the limits equal to ensure continuity

For \( f(x) \) to be continuous at \( x = 0 \):

\[ \lim_{x \to 0^-} f(x) = f(0) = \lim_{x \to 0^+} f(x) \]

Thus:

\[ 4 = a = 8 \]

### Conclusion

Since both limits must equal \( a \), we find that:

\[ a = 8 \]