https://file.aiquickdraw.com/mathbot/file/1726308412273l0rk7vmf.jpg

Let's analyze each limit expression to determine whether the limit exists and is a real number.

1. **First Expression:**
   \[
   \lim_{x \to \infty} \frac{\sin(8x) \cos(e^{x^2 + 9})}{x^8 - 9}
   \]
   - As \(x \to \infty\), \(x^8 - 9\) grows rapidly towards infinity.
   - The numerator involves \(\sin(8x)\), which oscillates between -1 and 1, and \(\cos(e^{x^2+9})\), which also oscillates.
   - However, due to the growth of \(x^8\) in the denominator, this expression tends to 0 because oscillations in the numerator are divided by a rapidly growing denominator.
   - Therefore, the limit exists and is 0, a real number.

2. **Second Expression:**
   \[
   \lim_{x \to 8} (x + 2) \sin\left(\frac{1}{x - 8}\right)
   \]
   - As \(x \to 8\), the term \(\frac{1}{x - 8}\) tends to infinity, causing \(\sin\left(\frac{1}{x - 8}\right)\) to oscillate infinitely.
   - Since the sine function oscillates wildly, the overall product also oscillates without tending to any specific value.
   - Therefore, this limit does not exist.

3. **Third Expression:**
   \[
   \lim_{x \to 6} \frac{x^2 + 7x - 78}{x - 6}
   \]
   - Direct substitution of \(x = 6\) gives \(0/0\), indicating that L'Hopital's Rule or factoring may be applied.
   - The expression can be factored as:
     \[
     \frac{(x - 6)(x + 13)}{x - 6}
     \]
   - Cancelling the common factor gives \(x + 13\), and substituting \(x = 6\) yields \(19\).
   - Therefore, the limit exists and is 19, a real number.

4. **Fourth Expression:**
   \[
   \lim_{x \to 13} \frac{4x - 13}{x - 13}
   \]
   - Direct substitution gives \(\frac{4(13) - 13}{13 - 13} = \frac{52 - 13}{0} = \frac{39}{0}\), which indicates a division by zero.
   - Since the numerator is non-zero and the denominator approaches zero, this is a case of infinite limit.
   - Hence, the limit does not exist as a real number.

### Conclusion:
The limits that exist and are real numbers are the first and third options, which have been correctly selected in the image.

Would you like more details on any of these solutions or any related questions?

Here are 5 related questions to deepen understanding:
1. How can L'Hopital's Rule be applied to solve indeterminate forms like \(0/0\)?
2. Why does the sine function cause certain limits to oscillate and not exist?
3. What happens to limits involving infinity in the denominator, like the first expression?
4. How can factoring help in solving limits with rational functions?
5. How do you determine when a limit approaches infinity versus when it does not exist?

**Tip:** When evaluating limits with \(0/0\) forms, always consider using factoring or L'Hopital's Rule for simplification.

Select ALL the options where the corresponding limit exists, i.e., the limit is a real number.

Learn how to determine if limits exist for complex functions, including those involving trigonometric expressions and indeterminate forms. This guide covers L'Hopital's Rule, factoring techniques, and the behavior of limits approaching infinity. Suitable for university-level calculus students.

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