It looks like you've uploaded an image of a limit problem that asks for the limit:

$\lim_{x \to 25} \frac{x - 25}{\sqrt{x + 5} - \sqrt{x - 5}}$

The hint suggests multiplying both the numerator and the denominator by $\sqrt{x + 5} + \sqrt{x - 5}$, which is a standard technique used to simplify expressions involving square roots.

Let's go through the steps again:

Step 1: Multiply by the conjugate

To eliminate the square roots in the denominator, multiply both the numerator and denominator by the conjugate of the denominator:

$\frac{x - 25}{\sqrt{x + 5} - \sqrt{x - 5}} \times \frac{\sqrt{x + 5} + \sqrt{x - 5}}{\sqrt{x + 5} + \sqrt{x - 5}}$

This gives:

$\frac{(x - 25)(\sqrt{x + 5} + \sqrt{x - 5})}{(\sqrt{x + 5})^2 - (\sqrt{x - 5})^2}$

Step 2: Simplify the denominator

Using the difference of squares formula, the denominator simplifies to:

$(\sqrt{x + 5})^2 - (\sqrt{x - 5})^2 = (x + 5) - (x - 5) = 10$

Step 3: Substitute $x = 25$

Now, the expression becomes:

$\frac{(x - 25)(\sqrt{x + 5} + \sqrt{x - 5})}{10}$

As $x \to 25$, we can directly substitute $x = 25$:

• $x - 25 \to 0$
• $\sqrt{x + 5} + \sqrt{x - 5} \to \sqrt{30} + \sqrt{20}$

Thus, the entire numerator goes to $0$, and the denominator remains $10$. So the limit is:

$\frac{0}{10} = 0$

Conclusion:

The limit is $0$.

Would you like further clarification or help with another problem? Here are five related questions to explore further:

1. How do you handle limits with square roots and indeterminate forms like $\frac{0}{0}$?
2. What is L'Hopital's Rule, and how can it be applied to limits like this one?
3. Can you use similar conjugate techniques for limits involving trigonometric functions?
4. What other algebraic manipulations can simplify limits in calculus?
5. How does the behavior of square roots impact the evaluation of limits as $x \to a$?

Tip: Always try to simplify the expression algebraically by factoring, multiplying by conjugates, or using other strategies before directly substituting the value. This can often resolve indeterminate forms.