We want to evaluate the limit:

$\lim_{x \to \infty} \frac{\sqrt{x^2 - 7}}{x - 3}$

Step 1: Simplify the expression inside the limit

Notice that as $x$ approaches infinity, $x^2$ dominates over 7, so $\sqrt{x^2 - 7}$ can be approximated by $\sqrt{x^2} = |x|$. Since $x$ is positive as $x$ approaches infinity, $|x| = x$. Therefore, we approximate:

$\sqrt{x^2 - 7} \approx x$

Step 2: Simplify the limit expression

Substituting the approximation into the limit:

$\lim_{x \to \infty} \frac{\sqrt{x^2 - 7}}{x - 3} \approx \lim_{x \to \infty} \frac{x}{x - 3}$

Step 3: Divide numerator and denominator by $x$

Simplify the expression by dividing both the numerator and the denominator by $x$:

$\lim_{x \to \infty} \frac{x}{x(1 - \frac{3}{x})} = \lim_{x \to \infty} \frac{1}{1 - \frac{3}{x}}$

Step 4: Evaluate the limit

As $x$ approaches infinity, $\frac{3}{x}$ approaches 0:

$\lim_{x \to \infty} \frac{1}{1 - 0} = \frac{1}{1} = 1$

Thus, the limit is:

$\boxed{1}$

Would you like further details or have any other questions?

Here are some related questions you might consider:

1. What happens if the limit is taken as $x$ approaches negative infinity?
2. How would the limit change if we replaced $-7$ with another constant?
3. Can you generalize this approach for any expression of the form $\sqrt{x^2 + C}$ over $x + D$?
4. What happens if you consider the limit where the denominator is $x + 3$ instead of $x - 3$?
5. How do you evaluate the limit if the square root term had a higher power of $x$ (e.g., $\sqrt{x^4 - 7}$)?

Tip: When dealing with limits that involve square roots of polynomials, it's often helpful to factor out the highest power of $x$ within the square root to simplify the expression.