To evaluate the limit:

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

First, let's analyze the form of the limit. When substituting $x = 7$ directly:

$\frac{\sqrt{7} - 7}{7 - 49} = \frac{7 - 7}{7 - 49} = \frac{0}{0}$

We get the indeterminate form $\frac{0}{0}$. Since this is indeterminate, we need to simplify the expression.

Step 1: Use Algebraic Manipulation (Rationalizing)

We can rationalize the numerator by multiplying both the numerator and denominator by the conjugate of the numerator:

$\frac{\sqrt{x} - 7}{x - 49} \cdot \frac{\sqrt{x} + 7}{\sqrt{x} + 7}$

This gives:

$\frac{(\sqrt{x} - 7)(\sqrt{x} + 7)}{(x - 49)(\sqrt{x} + 7)} = \frac{x - 49}{(x - 49)(\sqrt{x} + 7)}$

Step 2: Simplify the Expression

The $x - 49$ terms in the numerator and denominator cancel out, leaving:

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

Step 3: Evaluate the Limit

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

$\frac{1}{\sqrt{7} + 7} = \frac{1}{7 + 7} = \frac{1}{14}$

Thus, the value of the limit is:

$\lim_{{x \to 7}} \frac{\sqrt{x} - 7}{x - 49} = \frac{1}{14}$

Would you like further clarification on this process, or any additional details?

Here are 5 related questions for further exploration:

1. How do you rationalize expressions with different forms of radicals?
2. Can limits always be solved by rationalizing the numerator?
3. What is the significance of the indeterminate form $\frac{0}{0}$?
4. How does L'Hôpital's Rule apply to limits like this one?
5. How would the limit change if we had $\lim_{{x \to c}}$ where $c \neq 7$?

Tip: Always check for an indeterminate form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$ before proceeding with limit evaluation, as it indicates further algebraic manipulation is necessary.