The limit problem presented is:

$\lim_{x \to 7} \frac{2 - \sqrt{x - 3}}{x^2 - 49}$

Step 1: Simplify the denominator

The denominator is $x^2 - 49$, which is a difference of squares:

$x^2 - 49 = (x - 7)(x + 7)$

Step 2: Direct substitution check

If we directly substitute $x = 7$ into the expression:

• Numerator: $2 - \sqrt{7 - 3} = 2 - \sqrt{4} = 2 - 2 = 0$
• Denominator: $(7 - 7)(7 + 7) = 0$

This results in an indeterminate form $\frac{0}{0}$, so we must apply algebraic techniques to simplify further.

Step 3: Rationalize the numerator

To eliminate the square root in the numerator, multiply both the numerator and denominator by the conjugate of the numerator, $2 + \sqrt{x - 3}$:

$\frac{2 - \sqrt{x - 3}}{(x - 7)(x + 7)} \cdot \frac{2 + \sqrt{x - 3}}{2 + \sqrt{x - 3}} = \frac{(2 - \sqrt{x - 3})(2 + \sqrt{x - 3})}{(x - 7)(x + 7)(2 + \sqrt{x - 3})}$

The numerator becomes:

$(2 - \sqrt{x - 3})(2 + \sqrt{x - 3}) = 2^2 - (\sqrt{x - 3})^2 = 4 - (x - 3) = 4 - x + 3 = 7 - x$

Thus, the expression is simplified to:

$\frac{7 - x}{(x - 7)(x + 7)(2 + \sqrt{x - 3})}$

Step 4: Cancel common terms

Notice that $7 - x = -(x - 7)$, so we can cancel out the $(x - 7)$ term from the numerator and denominator:

$\frac{-1}{(x + 7)(2 + \sqrt{x - 3})}$

Step 5: Evaluate the limit

Now, substitute $x = 7$ into the simplified expression:

$\frac{-1}{(7 + 7)(2 + \sqrt{7 - 3})} = \frac{-1}{14(2 + \sqrt{4})} = \frac{-1}{14(2 + 2)} = \frac{-1}{14 \times 4} = \frac{-1}{56}$

Thus, the value of the limit is:

$\lim_{x \to 7} \frac{2 - \sqrt{x - 3}}{x^2 - 49} = \frac{-1}{56}$

Would you like any further details or explanations on this solution?

Here are 5 related questions to further explore:

1. How does the conjugate help simplify expressions with square roots?
2. Why do we cancel common terms in limits, and what happens if we don't?
3. What are the conditions that lead to indeterminate forms like $\frac{0}{0}$?
4. Can you apply a different method to evaluate limits besides algebraic simplification?
5. How do limits relate to derivatives in calculus?

Tip: Always check if you can factor expressions before using more advanced techniques like conjugates; this often simplifies the problem significantly.