`lim_(x to infty)[sqrt(x^(2) + ax + a^(2)) - sqrt(x^(2) + a^(2))]` is equal to `a/2`.

Answer

To solve the limit problem \( \lim_{x \to \infty} \left[ \sqrt{x^2 + ax + a^2} - \sqrt{x^2 + a^2} \right] \) and show that it equals \( \frac{a}{2} \), we will follow these steps:

### Step 1: Rationalization
We start by rationalizing the expression. We multiply and divide by the conjugate of the expression:

\[ L = \lim_{x \to \infty} \left[ \sqrt{x^2 + ax + a^2} - \sqrt{x^2 + a^2} \right] \cdot \frac{\sqrt{x^2 + ax + a^2} + \sqrt{x^2 + a^2}}{\sqrt{x^2 + ax + a^2} + \sqrt{x^2 + a^2}} \]

### Step 2: Simplifying the Numerator
Using the difference of squares, the numerator simplifies as follows:

\[ L = \lim_{x \to \infty} \frac{(x^2 + ax + a^2) - (x^2 + a^2)}{\sqrt{x^2 + ax + a^2} + \sqrt{x^2 + a^2}} \]

This simplifies to:

\[ L = \lim_{x \to \infty} \frac{ax}{\sqrt{x^2 + ax + a^2} + \sqrt{x^2 + a^2}} \]

### Step 3: Factor Out \( x^2 \) in the Denominator
Next, we will factor out \( x^2 \) from the square roots in the denominator:

\[ L = \lim_{x \to \infty} \frac{ax}{\sqrt{x^2(1 + \frac{a}{x} + \frac{a^2}{x^2})} + \sqrt{x^2(1 + \frac{a^2}{x^2})}} \]

This can be rewritten as:

\[ L = \lim_{x \to \infty} \frac{ax}{x\left(\sqrt{1 + \frac{a}{x} + \frac{a^2}{x^2}} + \sqrt{1 + \frac{a^2}{x^2}}\right)} \]

### Step 4: Cancel \( x \) in the Numerator and Denominator
Now, we can cancel \( x \) from the numerator and denominator:

\[ L = \lim_{x \to \infty} \frac{a}{\sqrt{1 + \frac{a}{x} + \frac{a^2}{x^2}} + \sqrt{1 + \frac{a^2}{x^2}}} \]

### Step 5: Evaluate the Limit
As \( x \to \infty \), \( \frac{a}{x} \) and \( \frac{a^2}{x^2} \) both approach 0. Thus, we have:

\[ L = \frac{a}{\sqrt{1 + 0 + 0} + \sqrt{1 + 0}} = \frac{a}{1 + 1} = \frac{a}{2} \]

### Conclusion
Thus, we have shown that:

\[ \lim_{x \to \infty} \left[ \sqrt{x^2 + ax + a^2} - \sqrt{x^2 + a^2} \right] = \frac{a}{2} \]