To solve the limit problem, let's go through the solution step-by-step.

The given problem is: $\lim_{{x \to \infty}} \left( \sqrt{x^2 - 6x} - x \right)$

Step 1: Simplify the Expression

To evaluate this limit, we can rationalize the expression by multiplying the numerator and the denominator by the conjugate, $\sqrt{x^2 - 6x} + x$.

So, rewrite the expression as: $\lim_{{x \to \infty}} \frac{\left( \sqrt{x^2 - 6x} - x \right) \cdot \left( \sqrt{x^2 - 6x} + x \right)}{\sqrt{x^2 - 6x} + x}$

Step 2: Expand the Numerator

Using the difference of squares, we get: $\lim_{{x \to \infty}} \frac{(x^2 - 6x) - x^2}{\sqrt{x^2 - 6x} + x}$ Simplify the numerator: $= \lim_{{x \to \infty}} \frac{x^2 - 6x - x^2}{\sqrt{x^2 - 6x} + x} = \lim_{{x \to \infty}} \frac{-6x}{\sqrt{x^2 - 6x} + x}$

Step 3: Factor Out $x$ in the Denominator

Now, we can factor $x$ out of the square root in the denominator: $= \lim_{{x \to \infty}} \frac{-6x}{x \left( \sqrt{1 - \frac{6}{x}} + 1 \right)}$ Cancel $x$ in the numerator and denominator: $= \lim_{{x \to \infty}} \frac{-6}{\sqrt{1 - \frac{6}{x}} + 1}$

Step 4: Take the Limit as $x \to \infty$

As $x$ approaches infinity, $\frac{6}{x} \to 0$. So, we get: $= \frac{-6}{\sqrt{1 - 0} + 1} = \frac{-6}{1 + 1} = \frac{-6}{2} = -3$

Conclusion

The answer is: $\boxed{-3}$

Would you like a further breakdown or explanation of any step?

Here are some related questions to deepen your understanding:

1. How would you solve this limit if the expression inside the square root were different, such as $x^2 + 6x$?
2. What if $x$ approached negative infinity instead of positive infinity?
3. Can you use L'Hôpital's Rule to evaluate this limit? If so, how would the steps differ?
4. What happens if you don’t rationalize the expression? Can you still solve the limit?
5. How would the solution change if the constant inside the square root was larger, e.g., $x^2 - 100x$?

Tip: Rationalizing expressions by multiplying with the conjugate is a useful strategy for limits that involve square roots.