The limit expression in the image is:

$\lim_{x \to \infty} \left( \sqrt{x^2 + 5x + 6} - \sqrt{x^2 + 7x - 6} \right)$

To evaluate this limit, we'll simplify the expression. Here's a step-by-step explanation:

Step 1: Rationalizing the Expression

The expression involves two square roots, which suggests using the technique of multiplying by the conjugate to eliminate the square roots. Multiply both the numerator and denominator by the conjugate of the expression inside the limit:

$\sqrt{x^2 + 5x + 6} - \sqrt{x^2 + 7x - 6} \quad \text{multiplied by} \quad \frac{\sqrt{x^2 + 5x + 6} + \sqrt{x^2 + 7x - 6}}{\sqrt{x^2 + 5x + 6} + \sqrt{x^2 + 7x - 6}}$

Step 2: Simplifying the Numerator

Using the difference of squares formula, the numerator becomes:

$(\sqrt{x^2 + 5x + 6})^2 - (\sqrt{x^2 + 7x - 6})^2 = (x^2 + 5x + 6) - (x^2 + 7x - 6)$

Simplifying the terms inside the parentheses:

$x^2 + 5x + 6 - (x^2 + 7x - 6) = x^2 + 5x + 6 - x^2 - 7x + 6 = -2x + 12$

So the numerator simplifies to $-2x + 12$.

Step 3: Simplifying the Denominator

The denominator is:

$\sqrt{x^2 + 5x + 6} + \sqrt{x^2 + 7x - 6}$

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

As $x$ approaches infinity, the highest power of $x$ dominates in both the numerator and denominator. So for large $x$, we approximate the square roots:

$\sqrt{x^2 + 5x + 6} \approx x \quad \text{and} \quad \sqrt{x^2 + 7x - 6} \approx x$

Thus, the denominator for large $x$ is approximately:

$\sqrt{x^2 + 5x + 6} + \sqrt{x^2 + 7x - 6} \approx x + x = 2x$

Step 5: Final Simplification

The limit expression now becomes:

$\lim_{x \to \infty} \frac{-2x + 12}{2x}$

Dividing both the numerator and denominator by $x$:

$\lim_{x \to \infty} \frac{-2 + \frac{12}{x}}{2}$

As $x \to \infty$, the term $\frac{12}{x} \to 0$, so the expression simplifies to:

$\frac{-2}{2} = -1$

Final Answer:

$\lim_{x \to \infty} \left( \sqrt{x^2 + 5x + 6} - \sqrt{x^2 + 7x - 6} \right) = -1$

Would you like further details on any steps or additional examples?

Here are five related questions to explore:

1. How would the result change if we had a different sign in front of the terms inside the square roots?
2. What happens if we use finite limits rather than limits approaching infinity?
3. How can this limit technique be generalized for other expressions involving square roots?
4. Can you solve limits at infinity involving cubic terms inside the square roots?
5. What is the geometric interpretation of subtracting two square root functions?

Tip: When dealing with limits involving square roots, multiplying by the conjugate is a very powerful tool for simplification!