Let's go through the limits one by one:

(a) $\lim_{x \to 3} (x^2 + 2x - 1)$

Since this function is a polynomial, we can directly substitute $x = 3$ into the function.

$x^2 + 2x - 1 = 3^2 + 2(3) - 1 = 9 + 6 - 1 = 14$

Answer:
$\lim_{x \to 3} (x^2 + 2x - 1) = 14$

(b) $\lim_{x \to 3} \frac{x^4 - 18x^2 + 81}{(x - 3)^2}$

Let's first simplify the numerator.

$x^4 - 18x^2 + 81 = (x^2 - 9)^2 = (x - 3)^2 (x + 3)^2$

Thus, the limit becomes:

$\lim_{x \to 3} \frac{(x - 3)^2 (x + 3)^2}{(x - 3)^2} = \lim_{x \to 3} (x + 3)^2$

Now, substituting $x = 3$:

$(3 + 3)^2 = 6^2 = 36$

Answer:
$\lim_{x \to 3} \frac{x^4 - 18x^2 + 81}{(x - 3)^2} = 36$

(c) $\lim_{t \to 2} \frac{\sqrt{(t + 4)(t - 2)^4}}{(3t - 6)^2}$

We start by simplifying the expressions.

• $3t - 6 = 3(t - 2)$, so $(3t - 6)^2 = 9(t - 2)^2$.
• The numerator is $\sqrt{(t + 4)(t - 2)^4} = (t - 2)^2 \sqrt{t + 4}$.

Thus, the limit becomes:

$\lim_{t \to 2} \frac{(t - 2)^2 \sqrt{t + 4}}{9(t - 2)^2} = \lim_{t \to 2} \frac{\sqrt{t + 4}}{9}$

Now, substituting $t = 2$:

$\frac{\sqrt{2 + 4}}{9} = \frac{\sqrt{6}}{9}$

Answer:
$\lim_{t \to 2} \frac{\sqrt{(t + 4)(t - 2)^4}}{(3t - 6)^2} = \frac{\sqrt{6}}{9}$

(d) $\lim_{u \to 2} \frac{(u + 2)(u^2 - u - 6)}{u^2 + 4u + 4}$

Let’s simplify the expressions.

• $u^2 - u - 6 = (u - 3)(u + 2)$
• $u^2 + 4u + 4 = (u + 2)^2$

Thus, the limit becomes:

$\lim_{u \to 2} \frac{(u + 2)(u - 3)(u + 2)}{(u + 2)^2} = \lim_{u \to 2} (u - 3)$

Now, substituting $u = 2$:

$2 - 3 = -1$

Answer:
$\lim_{u \to 2} \frac{(u + 2)(u^2 - u - 6)}{u^2 + 4u + 4} = -1$

Summary of Answers

1. (a) $14$
2. (b) $36$
3. (c) $\frac{\sqrt{6}}{9}$
4. (d) $-1$

Would you like further explanations or have any questions?
Here are 5 related questions to explore:

1. What are the types of indeterminate forms encountered in limits?
2. How do we apply L'Hôpital's rule to solve limits?
3. What is the difference between finite limits and infinite limits?
4. When do we need to factor expressions in limit problems?
5. What are the common techniques for solving square root limits?

Tip: Always try substituting the point of approach in limit problems first—if it doesn't yield an indeterminate form, you're done!