Let's analyze and solve each of the limit problems from the image one by one.

Problem (a)

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

Solution: We can directly substitute $x = 3$ into the expression, since it's a polynomial and does not have any discontinuities at $x = 3$. $\lim_{x \to 3} (3^2 + 2 \cdot 3 - 1) = 9 + 6 - 1 = 14$

Answer: $14$

Problem (b)

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

Solution: We need to simplify the numerator. Start by substituting $x = 3$ to check if the expression leads to an indeterminate form: $\text{Numerator: } 3^4 - 18 \cdot 3^2 + 81 = 81 - 162 + 81 = 0,$ $\text{Denominator: } (3 - 3)^2 = 0.$ Since both numerator and denominator are zero, it's an indeterminate form $\frac{0}{0}$. We should factor the numerator.

Factor the numerator: $x^4 - 18x^2 + 81 = (x^2 - 9)(x^2 - 9) = (x - 3)^2 (x + 3)^2.$ Now we can simplify: $\frac{(x - 3)^2 (x + 3)^2}{(x - 3)^2} = (x + 3)^2.$ $\lim_{x \to 3} (x + 3)^2 = (3 + 3)^2 = 6^2 = 36.$

Answer: $36$

Problem (c)

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

Solution: First, substitute $t = 2$ to check if the expression leads to an indeterminate form: $\text{Numerator: } \sqrt{(2 + 4)}(2 - 2)^4 = \sqrt{6} \cdot 0 = 0,$ $\text{Denominator: } (3 \cdot 2 - 6)^2 = 0^2 = 0.$ This is an indeterminate form $\frac{0}{0}$. We need to simplify.

Observe that: $3t - 6 = 3(t - 2).$ Thus, the denominator becomes: $(3(t - 2))^2 = 9(t - 2)^2.$ Simplify the limit: $\frac{\sqrt{(t + 4)}(t - 2)^4}{9(t - 2)^2} = \frac{\sqrt{(t + 4)}(t - 2)^2}{9}.$ Substitute $t = 2$: $\frac{\sqrt{(2 + 4)}(2 - 2)^2}{9} = \frac{\sqrt{6} \cdot 0}{9} = 0.$

Answer: $0$

Problem (d)

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

Solution: First, substitute $u = 2$ to check for indeterminate form: $\text{Numerator: } (2 + 2)(2^2 - 2 - 6) = 4 \cdot (4 - 2 - 6) = 4 \cdot 0 = 0,$ $\text{Denominator: } 2^2 + 4 \cdot 2 + 4 = 4 + 8 + 4 = 16.$ The numerator is $0$, and the denominator is $16$. Thus: $\lim_{u \to 2} \frac{0}{16} = 0.$

Answer: $0$

Summary of Answers:

1. (a) $14$
2. (b) $36$
3. (c) $0$
4. (d) $0$

Would you like further details on any of these solutions, or do you have any other questions?

Related Questions:

1. How do you factor higher-degree polynomials efficiently?
2. What are the common methods to handle indeterminate forms in limits?
3. How can you determine if a function is continuous at a point?
4. What are the conditions under which L'Hôpital's Rule can be applied?
5. How do square roots affect the simplification of rational expressions?

Tip:

Always check for indeterminate forms before proceeding with simplifications. If direct substitution results in $\frac{0}{0}$ or $\frac{\infty}{\infty}$, it signals the need for further simplification or special techniques like factoring or L'Hôpital's Rule.