To solve the limit

$\lim_{x \to 4} \left(\frac{3(x-4)\sqrt{x+5}}{3-\sqrt{x+5}}\right),$

we begin by substituting $x = 4$ into the expression to see if it results in an indeterminate form:

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

Since substituting directly gives an indeterminate form $\frac{0}{0}$, we need to simplify the expression further.

Simplification

The expression inside the limit is:

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

To eliminate the indeterminate form, we rationalize the denominator by multiplying both the numerator and the denominator by the conjugate of the denominator:

$\text{Conjugate of } 3 - \sqrt{x+5} \text{ is } 3 + \sqrt{x+5}.$

Multiply both the numerator and denominator by $3 + \sqrt{x+5}$:

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

Simplify the denominator using the difference of squares:

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

So, the limit becomes:

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

Notice that $4 - x = -(x - 4)$. Substituting this into the expression:

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

Now, we can cancel out $x - 4$ from the numerator and the denominator:

$\lim_{x \to 4} -3\sqrt{x+5}(3+\sqrt{x+5}).$

Finally, substitute $x = 4$ into the simplified expression:

$-3\sqrt{4+5}(3+\sqrt{4+5}) = -3\sqrt{9}(3+\sqrt{9}) = -3(3)(3+3) = -3 \times 3 \times 6 = -54.$

Final Answer

The limit is:

$\lim_{x \to 4} \left(\frac{3(x-4)\sqrt{x+5}}{3-\sqrt{x+5}}\right) = -54.$

Would you like more details or have any questions?

Here are some related questions for further exploration:

1. How do you rationalize a denominator?
2. What is the difference of squares and how is it used?
3. How can you identify indeterminate forms?
4. What other methods can be used to solve limits?
5. How does the conjugate help in simplifying expressions?
6. Can this problem be solved using L'Hôpital's Rule?
7. What are other common forms that result in indeterminate expressions?
8. How does factoring help in limit problems?

Tip: When you encounter a limit problem that leads to an indeterminate form, always consider algebraic techniques like rationalization, factoring, or using conjugates before applying more advanced methods like L'Hôpital's Rule.