https://file.aiquickdraw.com/mathbot/file/1729051805028buoqcyps.jpg

Let's go through each limit problem step-by-step.

---

### 1. \( \lim_{x \to 3} 7 \)
This is a constant function, so the limit will be the same as the constant value:

\[
\lim_{x \to 3} 7 = 7
\]

---

### 2. \( \lim_{x \to 1} (2x - 4x^2) \)

Direct substitution:

\[
2(1) - 4(1)^2 = 2 - 4 = -2
\]

---

### 3. \( \lim_{x \to 2} \frac{2x^2 + 4x + 4}{3x - 2} \)

Direct substitution first:

\[
\frac{2(2)^2 + 4(2) + 4}{3(2) - 2} = \frac{8 + 8 + 4}{6 - 2} = \frac{20}{4} = 5
\]

---

### 4. \( \lim_{x \to 4} \frac{\sqrt{x^2 + 9}}{2x} \)

Evaluate by direct substitution:

\[
\frac{\sqrt{4^2 + 9}}{2(4)} = \frac{\sqrt{16 + 9}}{8} = \frac{\sqrt{25}}{8} = \frac{5}{8}
\]

---

### 5. \( \lim_{x \to 1} \frac{x^2 + 4x - 5}{x - 1} \)

Direct substitution gives:

\[
\frac{1^2 + 4(1) - 5}{1 - 1} = \frac{0}{0}
\]

Since this is an indeterminate form, let's factor the numerator:

\[
x^2 + 4x - 5 = (x - 1)(x + 5)
\]

Now:

\[
\frac{(x - 1)(x + 5)}{x - 1} = x + 5 \quad \text{(for } x \neq 1)
\]

Now substitute \(x = 1\):

\[
1 + 5 = 6
\]

---

### 6. \( \lim_{x \to 2} \frac{x^2 + 5x + 6}{x^2 - 3x - 10} \)

Direct substitution gives:

\[
\frac{2^2 + 5(2) + 6}{2^2 - 3(2) - 10} = \frac{4 + 10 + 6}{4 - 6 - 10} = \frac{20}{-12} = -\frac{5}{3}
\]

---

### 7. \( \lim_{x \to 2} \frac{4 - x^2}{3 - \sqrt{x^2 + 5}} \)

Direct substitution gives:

\[
\frac{4 - 2^2}{3 - \sqrt{2^2 + 5}} = \frac{4 - 4}{3 - \sqrt{9}} = \frac{0}{0}
\]

This is an indeterminate form. To resolve it, multiply the numerator and denominator by the conjugate:

\[
\frac{4 - x^2}{3 - \sqrt{x^2 + 5}} \cdot \frac{3 + \sqrt{x^2 + 5}}{3 + \sqrt{x^2 + 5}}
\]

Now:

\[
\frac{(4 - x^2)(3 + \sqrt{x^2 + 5})}{9 - (x^2 + 5)}
\]

Since \(4 - x^2 = (2 - x)(2 + x)\), simplify the denominator:

\[
9 - (x^2 + 5) = 4 - x^2 = (2 - x)(2 + x)
\]

Now cancel out \((2 - x)(2 + x)\) from the numerator and denominator, and substitute \(x = 2\):

\[
3 + \sqrt{2^2 + 5} = 3 + \sqrt{9} = 3 + 3 = 6
\]

Thus:

\[
\lim_{x \to 2} \frac{4 - x^2}{3 - \sqrt{x^2 + 5}} = 6
\]

---

## Final Answers

1. \( 7 \)  
2. \( -2 \)  
3. \( 5 \)  
4. \( \frac{5}{8} \)  
5. \( 6 \)  
6. \( -\frac{5}{3} \)  
7. \( 6 \)  

---

Do you have any questions, or would you like further explanation on any of these problems?

---

### Related Questions
1. How do you handle indeterminate forms like \( \frac{0}{0} \) in limits?
2. What are common techniques for factoring quadratic expressions?
3. How does multiplying by the conjugate help in simplifying limits?
4. Can L'Hôpital's rule be applied to all \( \frac{0}{0} \) cases?
5. How can limits at infinity be handled differently from finite limits?

### Tip  
When direct substitution gives an indeterminate form, try simplifying the expression, factoring, or multiplying by the conjugate.

This guide shows how to compute limits of various functions using direct substitution, factoring, and rationalization. Problems involve indeterminate forms and algebraic manipulation. Suitable for high school students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation