The problem asks to evaluate the limit:

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

Step-by-Step Solution:

1. Direct Substitution:

   • Let's first try substituting $x = 2$ directly: $\frac{\sqrt{2 + 7} - 3}{2 - 2} = \frac{\sqrt{9} - 3}{0} = \frac{3 - 3}{0} = \frac{0}{0}$ This gives an indeterminate form $\frac{0}{0}$, so we need to simplify using another method.

2. Multiply by the Conjugate: To eliminate the square root, we can multiply both the numerator and denominator by the conjugate of the numerator: $\frac{\sqrt{x + 7} - 3}{x - 2} \times \frac{\sqrt{x + 7} + 3}{\sqrt{x + 7} + 3}$ This results in: $\frac{\left(\sqrt{x + 7} - 3\right)\left(\sqrt{x + 7} + 3\right)}{(x - 2)\left(\sqrt{x + 7} + 3\right)}$ Using the difference of squares in the numerator: $= \frac{(x + 7) - 9}{(x - 2)(\sqrt{x + 7} + 3)}$ Simplify the numerator: $= \frac{x - 2}{(x - 2)(\sqrt{x + 7} + 3)}$ Now, cancel out the common factor of $x - 2$ in the numerator and denominator: $= \frac{1}{\sqrt{x + 7} + 3}$

3. Substitute $x = 2$ again: Now, substitute $x = 2$ into the simplified expression: $= \frac{1}{\sqrt{2 + 7} + 3} = \frac{1}{\sqrt{9} + 3} = \frac{1}{3 + 3} = \frac{1}{6}$

Conclusion:

The value of the limit is $\frac{1}{6}$.

It appears the correct answer was marked wrong in the system, but based on this detailed solution, the correct answer is indeed $\frac{1}{6}$.

Would you like more details or have any questions?

Follow-up Questions:

1. Why do we multiply by the conjugate to simplify the expression?
2. What other methods can be used to evaluate limits that give an indeterminate form?
3. How do difference of squares work, and why does it help simplify?
4. Can we use L'Hôpital's Rule here, and how would it apply?
5. What are other common indeterminate forms and strategies for resolving them?

Tip:

Always check for algebraic simplifications or indeterminate forms before concluding that a limit does not exist.