Let `f(x) = {((1-cos 4x)/(x^(2))",",x lt 0),(a",",x = 0),((sqrt(x))/(sqrt(16+sqrt(x))-4))",",x gt 0):}` Then, the value of a if possible, so that the function is continuous at x = 0, is……….

Answer

To find 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 the right-hand limit at \( x = 0 \) are equal to \( f(0) \).

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

### Step 1: Calculate the left-hand limit as \( x \) approaches 0 from the left
We need to find:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{1 - \cos(4x)}{x^2} \]
Using the identity \( 1 - \cos(4x) = 2 \sin^2(2x) \), we can rewrite the limit:
\[ \lim_{x \to 0^-} \frac{1 - \cos(4x)}{x^2} = \lim_{x \to 0^-} \frac{2 \sin^2(2x)}{x^2} \]

### Step 2: Simplify the limit
We can use the substitution \( y = 2x \), which gives \( x = \frac{y}{2} \) and as \( x \to 0 \), \( y \to 0 \):
\[ \lim_{x \to 0^-} \frac{2 \sin^2(2x)}{x^2} = \lim_{y \to 0} \frac{2 \sin^2(y)}{(\frac{y}{2})^2} = \lim_{y \to 0} \frac{2 \sin^2(y)}{\frac{y^2}{4}} = \lim_{y \to 0} \frac{8 \sin^2(y)}{y^2} \]
Using the limit \( \lim_{y \to 0} \frac{\sin(y)}{y} = 1 \):
\[ \lim_{y \to 0} \frac{8 \sin^2(y)}{y^2} = 8 \cdot 1 = 8 \]

### Step 3: Calculate the right-hand limit as \( x \) approaches 0 from the right
Next, we find:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\sqrt{x}}{\sqrt{16 + \sqrt{x}} - 4} \]
To simplify this limit, we can multiply the numerator and the denominator by the conjugate of the denominator:
\[ \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{16 + \sqrt{x}} + 4)}{(\sqrt{16 + \sqrt{x}} - 4)(\sqrt{16 + \sqrt{x}} + 4)} = \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{16 + \sqrt{x}} + 4)}{16 + \sqrt{x} - 16} = \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{16 + \sqrt{x}} + 4)}{\sqrt{x}} \]
This simplifies to:
\[ \lim_{x \to 0^+} \sqrt{16 + \sqrt{x}} + 4 \]
As \( x \to 0 \), \( \sqrt{x} \to 0 \):
\[ \sqrt{16 + 0} + 4 = 4 + 4 = 8 \]

### Step 4: Set the limits equal to \( f(0) \)
For continuity at \( x = 0 \):
\[ \lim_{x \to 0^-} f(x) = f(0) = \lim_{x \to 0^+} f(x) \]
Thus:
\[ 8 = a = 8 \]

### Conclusion
The value of \( a \) that makes the function continuous at \( x = 0 \) is:
\[ \boxed{8} \]