The limit in the image can be written as:

$\lim_{x \to \infty} \left( \sqrt{9x^2 - 6x + 2} - (3x - 5) \right)$

We can simplify this step by step.

Step 1: Simplify the square root term

We begin by focusing on the square root expression $\sqrt{9x^2 - 6x + 2}$. As $x$ approaches infinity, the term $9x^2$ dominates because it grows much faster than the linear term $-6x$ or the constant $2$.

Thus, for large $x$, we approximate the expression inside the square root as: $\sqrt{9x^2 - 6x + 2} \approx \sqrt{9x^2} = 3x$

However, this approximation isn't exact, so we'll handle it rigorously by factoring $x^2$ from the terms inside the square root: $\sqrt{9x^2 - 6x + 2} = x\sqrt{9 - \frac{6}{x} + \frac{2}{x^2}}$

As $x \to \infty$, the fractions involving $x$ become very small, so: $\sqrt{9 - \frac{6}{x} + \frac{2}{x^2}} \to \sqrt{9} = 3$

Thus, for large $x$: $\sqrt{9x^2 - 6x + 2} \approx x \cdot 3 = 3x$

Step 2: Subtract the second term

Now subtract the second term from the expression: $3x - (3x - 5) = 3x - 3x + 5 = 5$

Step 3: Take the limit

Since the difference simplifies to $5$ as $x$ approaches infinity, we conclude: $\lim_{x \to \infty} \left( \sqrt{9x^2 - 6x + 2} - (3x - 5) \right) = 5$

Thus, the value of the limit is 5.

Let me know if you want more details, or have any questions!

Here are some related questions to practice:

1. What is the limit of $\lim_{x \to \infty} (\sqrt{x^2 + x + 1} - x)$?
2. How does the square root function behave as $x \to \infty$?
3. What would change in the solution if the expression under the square root had a higher-degree term?
4. How can L'Hôpital's Rule be used to find limits involving indeterminate forms?
5. What is the limit $\lim_{x \to \infty} (\sqrt{4x^2 - x} - 2x)$?

Tip: Always remember to simplify complex expressions by focusing on dominant terms for large values of $x$.