The problem shown in the image is:

$\lim_{x \to 4} \frac{\sqrt{x+5} - 3}{\sqrt{x - 4}}$

To solve this limit, let's follow these steps:

Step 1: Substitute $x = 4$ directly

Substituting $x = 4$ into the expression:

$\frac{\sqrt{4+5} - 3}{\sqrt{4 - 4}} = \frac{\sqrt{9} - 3}{\sqrt{0}} = \frac{3 - 3}{0} = \frac{0}{0}$

We get an indeterminate form $\frac{0}{0}$. So, we need to simplify the expression further.

Step 2: Rationalize the numerator

To resolve the indeterminate form, multiply the numerator and the denominator by the conjugate of the numerator:

$\frac{\sqrt{x+5} - 3}{\sqrt{x - 4}} \times \frac{\sqrt{x+5} + 3}{\sqrt{x+5} + 3} = \frac{(\sqrt{x+5} - 3)(\sqrt{x+5} + 3)}{(\sqrt{x - 4})(\sqrt{x+5} + 3)}$

Simplify the numerator using the difference of squares:

$(\sqrt{x+5} - 3)(\sqrt{x+5} + 3) = (x+5) - 9 = x - 4$

So the expression becomes:

$\frac{x - 4}{\sqrt{x - 4} \cdot (\sqrt{x+5} + 3)}$

Simplify further:

$= \frac{\sqrt{x - 4} \cdot \sqrt{x - 4}}{\sqrt{x - 4} \cdot (\sqrt{x+5} + 3)} = \frac{\sqrt{x - 4}}{\sqrt{x+5} + 3}$

Step 3: Cancel the common factor

$= \frac{\sqrt{x - 4}}{\sqrt{x+5} + 3}$

Now substitute $x = 4$:

$\frac{\sqrt{4 - 4}}{\sqrt{4+5} + 3} = \frac{\sqrt{0}}{\sqrt{9} + 3} = \frac{0}{3 + 3} = 0$

Final Answer:

$\lim_{x \to 4} \frac{\sqrt{x+5} - 3}{\sqrt{x - 4}} = 0$

Would you like more details or have any questions?

Related Questions:

1. How do you handle indeterminate forms in limits?
2. What are common strategies to simplify complex limits?
3. Why does rationalizing the numerator help in limit problems?
4. How does L'Hôpital's Rule apply to this problem?
5. What are other ways to approach solving limits involving square roots?

Tip: Always check for indeterminate forms like $\frac{0}{0}$ or $\frac{\infty}{\infty}$ to decide the appropriate simplification technique!