The given limit problem is:

$\lim_{{x \to 9}} \frac{\sqrt{7x - 62} - 1}{x - 9}$

Approach:

This is an indeterminate form of type $\frac{0}{0}$ when you directly substitute $x = 9$. To resolve this, we can use rationalization by multiplying both the numerator and the denominator by the conjugate of the numerator.

Step-by-step solution:

1. Multiply by the conjugate: Multiply both the numerator and the denominator by the conjugate of the numerator, $\sqrt{7x - 62} + 1$:

   $\frac{\sqrt{7x - 62} - 1}{x - 9} \cdot \frac{\sqrt{7x - 62} + 1}{\sqrt{7x - 62} + 1} = \frac{(\sqrt{7x - 62})^2 - 1^2}{(x - 9)(\sqrt{7x - 62} + 1)}$

2. Simplify the numerator: Use the difference of squares in the numerator:

   $(\sqrt{7x - 62})^2 - 1^2 = (7x - 62) - 1 = 7x - 63$

   So the expression becomes:

   $\frac{7x - 63}{(x - 9)(\sqrt{7x - 62} + 1)}$

3. Factor the numerator: Factor out $7$ from the numerator:

   $7(x - 9)$

   Now the expression is:

   $\frac{7(x - 9)}{(x - 9)(\sqrt{7x - 62} + 1)}$

4. Cancel out $(x - 9)$:

   Since $x \neq 9$, we can cancel $(x - 9)$ from both the numerator and the denominator:

   $\frac{7}{\sqrt{7x - 62} + 1}$

5. Substitute $x = 9$:

   Now substitute $x = 9$ into the simplified expression:

   $\frac{7}{\sqrt{7(9) - 62} + 1} = \frac{7}{\sqrt{63 - 62} + 1} = \frac{7}{\sqrt{1} + 1} = \frac{7}{1 + 1} = \frac{7}{2}$

Final Answer:

$\lim_{{x \to 9}} \frac{\sqrt{7x - 62} - 1}{x - 9} = \frac{7}{2}$

Would you like further details or have any other questions?

Here are 5 related questions:

1. How do you recognize the indeterminate form of a limit problem?
2. What are some other methods for solving limits besides rationalization?
3. What happens if you don't cancel common factors in a limit problem?
4. How does the conjugate method help in limit problems with square roots?
5. How can you verify a limit using numerical methods?

Tip: Always check for factorable expressions or common terms in the numerator and denominator to simplify limits easily.