Let's evaluate each limit step by step.

First Limit:

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

Step 1: Simplifying the expression.

We can simplify this by multiplying both the numerator and denominator by the conjugate of the denominator, i.e., $\sqrt{x} + \sqrt{7}$. This helps to eliminate the square roots in the denominator.

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

Since $(\sqrt{x})^2 = x$ and $(\sqrt{7})^2 = 7$, the denominator becomes:

$x - 7$

Thus, the expression simplifies to:

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

As long as $x \neq 7$, we can cancel out $x - 7$ from both the numerator and the denominator:

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

Step 2: Evaluating the limit.

Now that the expression is simplified, we can directly substitute $x = 7$:

$\lim_{x \to 7} (\sqrt{x} + \sqrt{7}) = \sqrt{7} + \sqrt{7} = 2\sqrt{7}$

Thus, the limit is:

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

Second Limit:

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

Step 1: Simplifying the expression.

We can use a known limit identity for this type of expression:

$\lim_{x \to a} \frac{\sqrt{x} - \sqrt{a}}{x - a} = \frac{1}{2\sqrt{a}}$

In this case, $a = 7$. So applying the identity directly, we get:

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

Final Answer:

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

Summary of Results:

• First limit: $2\sqrt{7}$
• Second limit: $\frac{1}{2\sqrt{7}}$

Would you like further clarification on any of the steps?

Related Questions:

1. How would you evaluate a limit involving higher powers of square roots?
2. What happens if the limit approaches a value where the function is undefined?
3. How does multiplying by the conjugate simplify radical expressions?
4. Can you apply the L'Hopital's Rule to either of these limits? How?
5. What other functions commonly benefit from multiplying by conjugates?

Tip: When you encounter square roots in a limit, multiplying by the conjugate is often a very effective simplification strategy!