https://file.aiquickdraw.com/mathbot/file/1726304070570eo4ww5ht.jpg

Let's evaluate the limits in the image one by one to determine if they exist.

### 1. \(\lim_{x \to \infty} \frac{\sin(8x) \cos(e^{x^2 + 9})}{x^8 - 9}\)
- **Numerator**: \(\sin(8x)\) oscillates between \(-1\) and \(1\), and \(\cos(e^{x^2 + 9})\) also oscillates between \(-1\) and \(1\), meaning the numerator is bounded.
- **Denominator**: \(x^8 - 9\) grows without bound as \(x \to \infty\), because the \(x^8\) term dominates.

Since the numerator is bounded and the denominator tends to infinity, the whole expression tends to \(0\).

- **Limit exists and is 0**.

### 2. \(\lim_{x \to 8} (x + 2) \sin\left(\frac{1}{x - 8}\right)\)
- As \(x \to 8\), \(\frac{1}{x - 8}\) tends to infinity, so \(\sin\left(\frac{1}{x - 8}\right)\) oscillates wildly between \(-1\) and \(1\).
- However, \((x + 2)\) tends to a constant value (10), so the product of \(10\) and an oscillating term does **not settle on any specific value**.

- **Limit does not exist** due to the oscillatory nature.

### 3. \(\lim_{x \to 6} \frac{x^2 + 7x - 78}{x - 6}\)
- We can try factoring the numerator to simplify. The quadratic \(x^2 + 7x - 78\) can be factored:
  \[
  x^2 + 7x - 78 = (x - 6)(x + 13)
  \]
- Now, the expression becomes:
  \[
  \frac{(x - 6)(x + 13)}{x - 6}
  \]
  For \(x \neq 6\), we can cancel \(x - 6\), and the limit becomes:
  \[
  \lim_{x \to 6} (x + 13) = 19
  \]

- **Limit exists and is 19**.

### 4. \(\lim_{x \to 13} \frac{4x - 13}{x - 13}\)
- At \(x = 13\), both the numerator and denominator vanish, so we can apply L'Hôpital's Rule (differentiating both the numerator and the denominator):
  - The derivative of \(4x - 13\) is 4.
  - The derivative of \(x - 13\) is 1.
- Thus, the limit becomes:
  \[
  \lim_{x \to 13} \frac{4}{1} = 4
  \]

- **Limit exists and is 4**.

### Final Answers:
- The limits **exist** for the following cases:
  - First expression: \(\lim_{x \to \infty} = 0\)
  - Third expression: \(\lim_{x \to 6} = 19\)
  - Fourth expression: \(\lim_{x \to 13} = 4\)

- The second expression does **not** have a limit.

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Would you like more details on any of the steps? Here are 5 related questions to further explore the topic:

1. What is L'Hôpital's Rule, and when can it be applied?
2. How can we handle limits with oscillatory functions like sine and cosine?
3. Can we always factor quadratic polynomials to simplify limits?
4. Why do some limits not exist due to oscillatory behavior?
5. What are indeterminate forms, and how do we resolve them in limits?

**Tip:** When evaluating limits that result in \( \frac{0}{0} \), always consider simplifying the expression or using L'Hôpital's Rule.

Select ALL the options where the corresponding limit exists, i.e. the limit is a real number. The options are: 
1. lim(x→∞) [sin(8x) cos(e^(x^2 + 9)) / (x^8 - 9)]
2. lim(x→8) [(x+2) sin(1 / (x-8))]
3. lim(x→6) [(x^2 + 7x - 78) / (x - 6)]
4. lim(x→13) [(4x - 13) / (x - 13)]

Learn how to evaluate limits involving oscillating sine functions, polynomials, and L'Hôpital's Rule. This problem covers key calculus concepts, including determining whether limits exist and resolving indeterminate forms. Suitable for students in Grades 11-12.

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