Given that `f(x)= {((1-cos4x)/(x^(2))",","if "x lt 0),(a",","if "x =0),((sqrt(x))/(sqrt(16+sqrt(x))-4)",","if "x gt 0):}`
If f(x) is continuous at x=0 find the value of a.

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 \( f(0) \).

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 \( x \to 0^- \)

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

Using the limit property:
\[ \lim_{x \to 0} \frac{1 - \cos(ax)}{(ax)^2} = \frac{1}{2} \quad \text{for any constant } a \]
we can rewrite our limit:
\[ \lim_{x \to 0^-} \frac{1 - \cos(4x)}{x^2} = \lim_{x \to 0} \frac{1 - \cos(4x)}{(4x)^2} \cdot 16 \]
This gives us:
\[ = 16 \cdot \frac{1}{2} = 8 \]

### Step 2: Calculate the Right-Hand Limit as \( x \to 0^+ \)

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 rationalize the denominator:
\[ \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{16 + \sqrt{x}} + 4)}{(\sqrt{16 + \sqrt{x}} - 4)(\sqrt{16 + \sqrt{x}} + 4)} = \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{16 + \sqrt{x}} + 4)}{16 + \sqrt{x} - 16} \]
This simplifies to:
\[ \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{16 + \sqrt{x}} + 4)}{\sqrt{x}} = \lim_{x \to 0^+} (\sqrt{16 + \sqrt{x}} + 4) \]
As \( x \to 0 \):
\[ \sqrt{16 + \sqrt{x}} \to \sqrt{16} = 4 \]
Thus:
\[ \lim_{x \to 0^+} (\sqrt{16 + \sqrt{x}} + 4) = 4 + 4 = 8 \]

### Step 3: Set the Limits Equal to Each Other

For continuity at \( x = 0 \):
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0) \]
This gives us:
\[ 8 = a \]

### Conclusion

Therefore, the value of \( a \) that makes \( f(x) \) continuous at \( x = 0 \) is:
\[ \boxed{8} \]