https://file.aiquickdraw.com/mathbot/file/172699057316999xykdvm.jpg

The image you uploaded contains eight limit problems. I will go ahead and solve each of them below:

### 1. \(\lim_{x \to 2} (3x + 2)\)
We can directly substitute \(x = 2\) into the expression:
\[
\lim_{x \to 2} (3x + 2) = 3(2) + 2 = 6 + 2 = 8
\]
Answer: \( \boxed{8} \)

### 2. \(\lim_{x \to 2} (x^2 - 1)\)
Substituting \(x = 2\):
\[
\lim_{x \to 2} (x^2 - 1) = (2^2 - 1) = 4 - 1 = 3
\]
Answer: \( \boxed{3} \)

### 3. \(\lim_{x \to 0} \left( 1 + \frac{1}{x} \right)\)
As \(x \to 0\), the term \(\frac{1}{x}\) grows without bound, leading to:
\[
\lim_{x \to 0} \left( 1 + \frac{1}{x} \right) = \infty
\]
Answer: \( \boxed{\infty} \) (undefined as \(x\) approaches zero from the right)

### 4. \(\lim_{x \to 5} \sqrt{x - 1}\)
Substituting \(x = 5\):
\[
\lim_{x \to 5} \sqrt{x - 1} = \sqrt{5 - 1} = \sqrt{4} = 2
\]
Answer: \( \boxed{2} \)

### 5. \(\lim_{x \to 1} \frac{x^2 - 1}{x - 1}\)
This is an indeterminate form \( \frac{0}{0} \), so we factor the numerator:
\[
\frac{x^2 - 1}{x - 1} = \frac{(x - 1)(x + 1)}{x - 1}
\]
Canceling out the common factor:
\[
\lim_{x \to 1} (x + 1) = 1 + 1 = 2
\]
Answer: \( \boxed{2} \)

### 6. \(\lim_{x \to 0} \frac{x^2 - 3x}{x}\)
Factoring the numerator:
\[
\frac{x^2 - 3x}{x} = \frac{x(x - 3)}{x}
\]
Canceling \(x\) (for \(x \neq 0\)):
\[
\lim_{x \to 0} (x - 3) = 0 - 3 = -3
\]
Answer: \( \boxed{-3} \)

### 7. \(\lim_{x \to 3} \frac{|x - 3|}{x - 3}\)
This limit depends on the direction of \(x\) approaching 3:

- As \(x \to 3^+\) (from the right), \(|x - 3| = x - 3\), so:
  \[
  \lim_{x \to 3^+} \frac{|x - 3|}{x - 3} = \frac{x - 3}{x - 3} = 1
  \]

- As \(x \to 3^-\) (from the left), \(|x - 3| = -(x - 3)\), so:
  \[
  \lim_{x \to 3^-} \frac{|x - 3|}{x - 3} = \frac{-(x - 3)}{x - 3} = -1
  \]

Since the limits from the left and right are not equal, the limit does not exist.

Answer: \( \boxed{\text{DNE (Does Not Exist)}} \)

### 8. \(\lim_{x \to 0} \frac{|x| - x}{x}\)
We evaluate the limit by considering the cases:

- As \(x \to 0^+\) (from the right), \(|x| = x\), so:
  \[
  \lim_{x \to 0^+} \frac{|x| - x}{x} = \frac{x - x}{x} = 0
  \]

- As \(x \to 0^-\) (from the left), \(|x| = -x\), so:
  \[
  \lim_{x \to 0^-} \frac{|x| - x}{x} = \frac{-x - x}{x} = -2
  \]

Since the limits from the left and right are not equal, the limit does not exist.

Answer: \( \boxed{\text{DNE (Does Not Exist)}} \)

---

Would you like further details on any of these problems? I can also draw graphs for the limits if needed.

Here are five additional questions to extend your understanding:
1. What happens to a function's behavior when its limit does not exist at a point?
2. How do piecewise functions affect limit calculations?
3. What are the conditions for a limit to exist at a certain point?
4. How can we use factoring to simplify expressions when calculating limits?
5. What role does one-sided limits play in determining the behavior of functions?

**Tip**: Always check for indeterminate forms like \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \), which often require algebraic manipulation or L'Hopital's Rule.

Find the limits of the sets presented in the image and draw the graph.

Solve various limit problems step by step. This exercise covers limits involving direct substitution, factoring, and one-sided limits, with examples like lim(x->2)(3x+2) and lim(x->3) |x-3|/(x-3). Suitable for students in Grades 11-12 or early college.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation