https://r2.erweima.ai/mathbot/image/1727106232583qua71ads.jpg

puedes calcular si es que existen los límites de la imagen?

Let's now proceed to calculate the limits given in the image.

### (c) \( \lim_{n \to \infty} \left( \frac{\sqrt{2n}}{\sqrt{n + 2}} + \frac{\sqrt{6n - 4}}{\sqrt{3n - 5}} \right) \)

1. As \( n \to \infty \), for both terms inside the square roots, the behavior of the functions simplifies due to the dominant term \( n \) in both the numerator and denominator:
   \[
   \frac{\sqrt{2n}}{\sqrt{n+2}} \approx \frac{\sqrt{2n}}{\sqrt{n}} = \sqrt{2}
   \]
   and
   \[
   \frac{\sqrt{6n - 4}}{\sqrt{3n - 5}} \approx \frac{\sqrt{6n}}{\sqrt{3n}} = \sqrt{2}.
   \]
2. Adding the two limits together:
   \[
   \sqrt{2} + \sqrt{2} = 2\sqrt{2}.
   \]
Thus, the limit for (c) is:
\[
\boxed{2\sqrt{2}}.
\]

---

### (d) \( \lim_{n \to \infty} \left( \sqrt{\frac{5n^3 - 6n^2 - 2}{6n^6 - 2n - 2}} + \frac{1}{2 \cos\left(\frac{1}{n}\right)} \right) \)

1. For the first term:
   \[
   \frac{5n^3 - 6n^2 - 2}{6n^6 - 2n - 2} \sim \frac{n^3}{n^6} = \frac{1}{n^3}.
   \]
   As \( n \to \infty \), \( \frac{1}{n^3} \to 0 \), so the square root of this expression tends to 0:
   \[
   \sqrt{\frac{1}{n^3}} \to 0.
   \]
2. For the second term, as \( n \to \infty \):
   \[
   \frac{1}{2 \cos\left(\frac{1}{n}\right)} \to \frac{1}{2 \cos(0)} = \frac{1}{2}.
   \]
Thus, the limit for (d) is:
\[
\boxed{\frac{1}{2}}.
\]

---

### (e) \( \lim_{x \to \infty} \frac{5 - 6x^3}{\sqrt{x^6 + 7}} \)

1. The dominant term in the numerator is \( -6x^3 \), and in the denominator, the dominant term is \( \sqrt{x^6} = x^3 \).
2. Simplifying the fraction as \( x \to \infty \):
   \[
   \frac{5 - 6x^3}{\sqrt{x^6 + 7}} \approx \frac{-6x^3}{x^3} = -6.
   \]
Thus, the limit for (e) is:
\[
\boxed{-6}.
\]

Let me know if you'd like a detailed explanation or any further questions! Here are 5 related questions:

1. What methods can be used to evaluate limits involving square roots?
2. How do you handle limits of rational functions as \( n \to \infty \)?
3. Why does the dominant term in polynomials determine the behavior of limits?
4. What is the role of trigonometric functions in limits when the argument tends to zero or infinity?
5. Can limits approach infinity instead of a finite number?

**Tip**: When dealing with limits at infinity, focus on the highest degree term in the numerator and denominator to simplify your calculations!

Solve three challenging limit problems involving square roots and rational functions as the variables approach infinity. This detailed explanation covers key concepts such as dominant term behavior and limits involving trigonometric functions. Suitable for undergraduate 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