https://file.aiquickdraw.com/mathbot/file/1729261434094mdkoogwk.jpg

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.

1. Tentukan nilai dari limit berikut:
   a. lim(x→3) (x^2 + 2x - 1)
   b. lim(x→3) (x^4 - 18x^2 + 81) / (x - 3)^2
   c. lim(t→2) sqrt(t+4)(t - 2)^4 / (3t - 6)^2
   d. lim(u→2) (u+2)(u^2 - u - 6) / (u^2 + 4u + 4)

Learn how to solve limit problems involving polynomials and rational expressions. This includes factoring higher-degree polynomials and handling indeterminate forms. Suitable for students in Grades 11-12.

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