Math Problem Statement

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)]

Solution

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

1. limxsin(8x)cos(ex2+9)x89\lim_{x \to \infty} \frac{\sin(8x) \cos(e^{x^2 + 9})}{x^8 - 9}

  • Numerator: sin(8x)\sin(8x) oscillates between 1-1 and 11, and cos(ex2+9)\cos(e^{x^2 + 9}) also oscillates between 1-1 and 11, meaning the numerator is bounded.
  • Denominator: x89x^8 - 9 grows without bound as xx \to \infty, because the x8x^8 term dominates.

Since the numerator is bounded and the denominator tends to infinity, the whole expression tends to 00.

  • Limit exists and is 0.

2. limx8(x+2)sin(1x8)\lim_{x \to 8} (x + 2) \sin\left(\frac{1}{x - 8}\right)

  • As x8x \to 8, 1x8\frac{1}{x - 8} tends to infinity, so sin(1x8)\sin\left(\frac{1}{x - 8}\right) oscillates wildly between 1-1 and 11.

  • However, (x+2)(x + 2) tends to a constant value (10), so the product of 1010 and an oscillating term does not settle on any specific value.

  • Limit does not exist due to the oscillatory nature.

3. limx6x2+7x78x6\lim_{x \to 6} \frac{x^2 + 7x - 78}{x - 6}

  • We can try factoring the numerator to simplify. The quadratic x2+7x78x^2 + 7x - 78 can be factored: x2+7x78=(x6)(x+13)x^2 + 7x - 78 = (x - 6)(x + 13)

  • Now, the expression becomes: (x6)(x+13)x6\frac{(x - 6)(x + 13)}{x - 6} For x6x \neq 6, we can cancel x6x - 6, and the limit becomes: limx6(x+13)=19\lim_{x \to 6} (x + 13) = 19

  • Limit exists and is 19.

4. limx134x13x13\lim_{x \to 13} \frac{4x - 13}{x - 13}

  • At x=13x = 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 4x134x - 13 is 4.
    • The derivative of x13x - 13 is 1.
  • Thus, the limit becomes: limx1341=4\lim_{x \to 13} \frac{4}{1} = 4

  • Limit exists and is 4.

Final Answers:

  • The limits exist for the following cases:

    • First expression: limx=0\lim_{x \to \infty} = 0
    • Third expression: limx6=19\lim_{x \to 6} = 19
    • Fourth expression: limx13=4\lim_{x \to 13} = 4
  • The second expression does not have a limit.


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 00\frac{0}{0}, always consider simplifying the expression or using L'Hôpital's Rule.

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Math Problem Analysis

Mathematical Concepts

Limits
L'Hôpital's Rule
Oscillatory Behavior in Limits
Factoring Polynomials

Formulas

L'Hôpital's Rule: lim(x→a) f(x)/g(x) = lim(x→a) f'(x)/g'(x) when 0/0 or ∞/∞
Quadratic Factoring: x^2 + bx + c = (x - r1)(x - r2)

Theorems

L'Hôpital's Rule
Limit Properties
Factoring Quadratic Expressions

Suitable Grade Level

Grades 11-12, Calculus