The value of `lim_(x to oo)(sqrt(3x^(2)+sqrt(3x^(2)+sqrt(3x^(2))))-sqrt(3x^(2)))` is ___________.

Answer

To solve the limit

\[ \lim_{x \to \infty} \left( \sqrt{3x^2 + \sqrt{3x^2 + \sqrt{3x^2}}} - \sqrt{3x^2} \right), \]

we will follow these steps:

### Step 1: Rewrite the limit expression

We start with the limit expression:

\[ \lim_{x \to \infty} \left( \sqrt{3x^2 + \sqrt{3x^2 + \sqrt{3x^2}}} - \sqrt{3x^2} \right). \]

### Step 2: Rationalize the expression

To simplify this expression, we will rationalize it. We multiply and divide by the conjugate:

\[ \lim_{x \to \infty} \frac{\left( \sqrt{3x^2 + \sqrt{3x^2 + \sqrt{3x^2}}} - \sqrt{3x^2} \right) \left( \sqrt{3x^2 + \sqrt{3x^2 + \sqrt{3x^2}}} + \sqrt{3x^2} \right)}{\sqrt{3x^2 + \sqrt{3x^2 + \sqrt{3x^2}}} + \sqrt{3x^2}}. \]

### Step 3: Simplify the numerator

The numerator simplifies as follows:

\[ \left( \sqrt{3x^2 + \sqrt{3x^2 + \sqrt{3x^2}}} \right)^2 - \left( \sqrt{3x^2} \right)^2 = \left(3x^2 + \sqrt{3x^2 + \sqrt{3x^2}} - 3x^2\right) = \sqrt{3x^2 + \sqrt{3x^2}}. \]

### Step 4: Substitute back into the limit

Now, we have:

\[ \lim_{x \to \infty} \frac{\sqrt{3x^2 + \sqrt{3x^2}}}{\sqrt{3x^2 + \sqrt{3x^2 + \sqrt{3x^2}}} + \sqrt{3x^2}}. \]

### Step 5: Factor out \(x\)

Next, we factor out \(x^2\) from the square roots:

\[ \sqrt{3x^2 + \sqrt{3x^2}} = x \sqrt{3 + \frac{\sqrt{3x^2}}{x^2}} = x \sqrt{3 + \frac{\sqrt{3}}{x}}. \]

As \(x \to \infty\), \(\frac{\sqrt{3}}{x} \to 0\), so:

\[ \sqrt{3 + \frac{\sqrt{3}}{x}} \to \sqrt{3}. \]

### Step 6: Evaluate the limit

Now substituting back, we get:

\[ \lim_{x \to \infty} \frac{x \sqrt{3}}{x \left( \sqrt{3} + \sqrt{3} \right)} = \lim_{x \to \infty} \frac{\sqrt{3}}{2\sqrt{3}} = \frac{1}{2}. \]

### Final Answer

Thus, the value of the limit is:

\[ \frac{1}{2}. \]