https://file.aiquickdraw.com/mathbot/file/17258442231338zqjudrz.jpg

The image contains four limit problems, and it asks to estimate the value of the limit using a table of values or to confirm the result graphically. I'll go ahead and solve each of these limit problems:

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

This is an indeterminate form of \(\frac{0}{0}\) when directly substituting \(x = 2\). To solve:

Factor the numerator:
\[
\frac{x^2 - 4}{x - 2} = \frac{(x - 2)(x + 2)}{x - 2}
\]

Cancel the common factor:
\[
\lim_{x \to 2} \frac{(x - 2)(x + 2)}{x - 2} = \lim_{x \to 2} (x + 2)
\]

Now, substitute \(x = 2\):
\[
\lim_{x \to 2} (x + 2) = 4
\]

**Answer: \(\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = 4\)**

---

### 25. \(\lim_{x \to 0} \frac{\sin 3x}{x}\)

This limit is a standard trigonometric limit, but we can rewrite it using the known limit \(\lim_{u \to 0} \frac{\sin u}{u} = 1\):

\[
\lim_{x \to 0} \frac{\sin 3x}{x} = \lim_{x \to 0} \frac{\sin 3x}{3x} \cdot 3 = 3 \cdot \lim_{x \to 0} \frac{\sin 3x}{3x} = 3 \cdot 1 = 3
\]

**Answer: \(\lim_{x \to 0} \frac{\sin 3x}{x} = 3\)**

---

### 26. \(\lim_{x \to 1} \frac{1}{x-1}\)

This limit is a standard form that tends to infinity as \(x\) approaches 1. The behavior of the limit depends on the direction from which \(x\) approaches 1:

- If \(x \to 1^+\), then \(\frac{1}{x-1}\) tends to \(+\infty\).
- If \(x \to 1^-\), then \(\frac{1}{x-1}\) tends to \(-\infty\).

Since the limit is not specified as one-sided, the two-sided limit does not exist because the left-hand limit and the right-hand limit are not equal.

**Answer: The limit does not exist (DNE)**

---

### 27. \(\lim_{x \to 1} \frac{9^x - 5^x}{x-1}\)

This limit can be solved using L'Hôpital's Rule, as it is in an indeterminate form \(\frac{0}{0}\).

First, differentiate the numerator and denominator:

- The derivative of the numerator \(9^x - 5^x\) is \(\ln(9) \cdot 9^x - \ln(5) \cdot 5^x\).
- The derivative of the denominator \(x-1\) is 1.

Thus, applying L'Hôpital's Rule:

\[
\lim_{x \to 1} \frac{9^x - 5^x}{x-1} = \lim_{x \to 1} \frac{\ln(9) \cdot 9^x - \ln(5) \cdot 5^x}{1} = \ln(9) \cdot 9^1 - \ln(5) \cdot 5^1
\]

Calculate this:

\[
\ln(9) \cdot 9 - \ln(5) \cdot 5 = 9\ln(9) - 5\ln(5)
\]

**Answer: \(\lim_{x \to 1} \frac{9^x - 5^x}{x-1} = 9\ln(9) - 5\ln(5)\)**

---

Would you like any further details on these solutions? 

Here are some relative questions to explore:

1. How do you determine when to apply L'Hôpital's Rule?
2. What are other methods for solving limits that result in indeterminate forms?
3. What are some common trigonometric limits used in calculus?
4. How do one-sided limits differ from two-sided limits?
5. What is the significance of factoring in solving limit problems?

**Tip:** When solving limits, always check if direct substitution leads to an indeterminate form, as this often indicates the need for factoring, simplification, or using L'Hôpital's Rule.

Solve the limits: lim (x -> 2) (x^2 - 4) / (x - 2), lim (x -> 0) sin(3x) / x, lim (x -> 1) 1 / (x - 1), lim (x -> 1) (9^x - 5^x) / (x - 1).

This page provides a step-by-step solution to limit problems, including rational functions and trigonometric limits. It covers concepts such as factoring, L'Hopital's Rule, and trigonometric identities. Suitable for high school or introductory calculus 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