Math Problem Statement

Выберите правильные утверждения: анализ утверждений о функции f(x) и g(x) и их порядках малости, эквивалентности sin и arcsin, неравенства пределов, и логарифмической аппроксимации для малых величин.

Solution

To analyze the statements in this image and confirm which ones are correct, let's go through each option in detail. Here’s a breakdown of each statement in the image:

  1. Statement 1:
    This statement talks about two functions f(x)f(x) and g(x)g(x) defined in some neighborhood of point aa with limits: limxaf(x)=0andlimxag(x)=0.\lim_{x \to a} f(x) = 0 \quad \text{and} \quad \lim_{x \to a} g(x) = 0. It then mentions that if limxaf(x)g(x)=C0\lim_{x \to a} \frac{f(x)}{g(x)} = C \neq 0, then f(x)f(x) and g(x)g(x) are infinitesimal functions of the same order of smallness as xax \to a.
    This is a true statement. When two functions both tend to zero and their ratio tends to a non-zero constant, they are of the same order of infinitesimality as xax \to a.

  2. Statement 2:
    This statement assumes α(x)\alpha(x) is an infinitesimally small quantity as xx0x \to x_0 (where x0x_0 \to \infty). It then claims: sinαarcsinαasxx0.\sin \alpha \sim \arcsin \alpha \quad \text{as} \quad x \to x_0. This is also true. For small values of α\alpha, sinαarcsinα\sin \alpha \approx \arcsin \alpha, as they behave similarly in terms of asymptotics.

  3. Statement 3:
    This statement considers two functions f(x)f(x) and g(x)g(x) in a neighborhood of point aa, with: f(x)g(x).f(x) \leq g(x). It states that if both functions have limits as xax \to a, then limxaf(x)limxag(x)\lim_{x \to a} f(x) \leq \lim_{x \to a} g(x).
    This is indeed correct and aligns with the properties of limits. If f(x)g(x)f(x) \leq g(x) in some neighborhood of aa, then their limits must satisfy limxaf(x)limxag(x)\lim_{x \to a} f(x) \leq \lim_{x \to a} g(x).

  4. Statement 4:
    This statement assumes α(x)\alpha(x) is an infinitesimally small quantity as xx0x \to x_0. It claims: loga(1+α)=αlnaasxx0.\log_a (1 + \alpha) = \frac{\alpha}{\ln a} \quad \text{as} \quad x \to x_0. This is also correct. For small α\alpha, loga(1+α)αlna\log_a(1 + \alpha) \approx \frac{\alpha}{\ln a} due to the first-order Taylor expansion of the logarithm function.

Conclusion

Based on the detailed analysis, all four statements appear to be correct. Since the last statement is already selected in the image, it suggests that this answer was correctly identified.

Would you like further details on any of these points, or do you have any specific questions?

Related Questions

  1. What does it mean for two functions to be of "the same order of smallness" as xax \to a?
  2. How is the asymptotic equivalence between sinα\sin \alpha and arcsinα\arcsin \alpha derived for small α\alpha?
  3. How does the comparison theorem for limits apply in cases with inequalities between functions?
  4. Why is the approximation loga(1+α)αlna\log_a (1 + \alpha) \approx \frac{\alpha}{\ln a} valid for small α\alpha?
  5. Can you give examples of functions that exhibit similar behavior to the ones in these statements?

Tip

When working with asymptotics, Taylor series expansions often provide powerful approximations for functions near specific points.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Infinitesimal Analysis
Asymptotic Behavior
Limits
Taylor Expansion

Formulas

lim(x→a) f(x) = 0
lim(x→a) g(x) = 0
lim(x→a) (f(x)/g(x)) = C ≠ 0
sin α ≈ arcsin α for small α
log_a(1 + α) ≈ α / ln(a) for small α

Theorems

Limit Comparison Theorem
Taylor Series Expansion

Suitable Grade Level

Undergraduate Mathematics