Question

Let S be the set of all \((α, β) ∈ \R\times\R\) such that
\(\lim\limits_{x\rightarrow\infin}\frac{\sin(x^2)(\log_ex)^\alpha\sin(\frac{1}{x^2})}{x^{αβ}(\log_e(1+x))^β}=0\)
Then which of the following is (are) correct ?

• A. (-1, 3) ∈ S
• B. (-1, 1) ∈ S
• C. (1, -1) ∈ S
• D. (1, -2) ∈ S

Answer 1

The Correct Option is B, C

1. Analyzing the Given Limit:
The given problem involves finding the limit of the following expression as $ x \to \infty $:

$ \lim_{x \to \infty} \frac{\sin(x^2) \cdot \sin\left(\frac{1}{x^2} \ln x \right)^\alpha}{x^\beta \left( \ln(1 + x) \right)^\beta} = 0 $

2. Breaking Down the Expression:
We begin by simplifying the expression as follows:

$ = \lim_{x \to \infty} \frac{\left( \sin(x^2) \right) \cdot \left( \frac{1}{x^2} \ln x \right)^\alpha}{x^\beta \left( \ln(1 + x) \right)^\beta} = 0 $

3. Analyzing the Behavior of $\sin(x^2)$:
The function $ \sin(x^2) $ oscillates between -1 and 1, so we can focus on the remaining terms.

4. Simplifying the Terms:
As $ x \to \infty $, we know that:

• $ \frac{1}{x^2} \to 0 $, so $ \sin\left(\frac{1}{x^2} \ln x \right) \approx \frac{1}{x^2} \ln x $.
• $ \ln(1 + x) \sim \ln x $ as $ x \to \infty $.

5. Final Expression for the Limit:
Using the approximations, the limit becomes:

$ = \lim_{x \to \infty} \frac{\left( \frac{1}{x^2} \ln x \right)^\alpha}{x^\beta \left( \ln x \right)^\beta} $

6. Simplifying Further:
We can now simplify the expression further:

$ = \lim_{x \to \infty} \frac{\ln^\alpha x}{x^{2\alpha + \beta} \ln^\beta x} $

$ = \lim_{x \to \infty} \frac{1}{x^{2\alpha + \beta} \ln^{\beta - \alpha} x} $

7. Condition for the Limit to Tend to Zero:
For this expression to tend to zero, the power of $x$ must be greater than zero, so we require:

$ 2\alpha + \beta > 0 $

8. Condition for $\alpha \beta + 2 > 0$:
We are given that it is possible if:

$ \alpha \beta + 2 > 0 \Rightarrow \alpha \beta > -2 $

9. Conclusion:
The condition is satisfied for the following values of $ \alpha \beta $:

Final Answer:
The correct option is: Option B and C

Answer 2

To solve the problem, we analyze the limit:

\[ \lim_{x \to \infty} \frac{\sin(x^2)(\log_e x)^\alpha \sin\left(\frac{1}{x^2}\right)}{x^{\alpha \beta} (\log_e(1+x))^\beta} = 0 \] for \((\alpha, \beta) \in \mathbb{R} \times \mathbb{R}\).

 

1. Behavior of terms as \(x \to \infty\):
- \(\sin(x^2)\) oscillates between \(-1\) and \(1\), bounded.
- \(\sin\left(\frac{1}{x^2}\right) \approx \frac{1}{x^2}\) for large \(x\).
- \(\log_e(1+x) \sim \log_e x\) for large \(x\).
Rewrite the limit as:

\[ \lim_{x \to \infty} \frac{\sin(x^2) (\log x)^\alpha \times \frac{1}{x^2}}{x^{\alpha \beta} (\log x)^\beta} = \lim_{x \to \infty} \sin(x^2) \frac{(\log x)^{\alpha - \beta}}{x^{2 + \alpha \beta}} \]

2. Since \(\sin(x^2)\) is bounded, consider the absolute value:

\[ \left|\frac{(\log x)^{\alpha - \beta}}{x^{2 + \alpha \beta}}\right| \]

3. For the limit to be zero, denominator must dominate numerator:
\[ x^{2 + \alpha \beta} \to \infty \quad \text{faster than} \quad (\log x)^{\alpha - \beta} \] As \(x \to \infty\), any positive power of \(x\) dominates any power of \(\log x\). So, the limit is zero if and only if:

\[ 2 + \alpha \beta > 0 \]

4. Check each given point \((\alpha, \beta)\):

• \((-1, 3)\):
  \[ 2 + (-1) \times 3 = 2 - 3 = -1 < 0 \] Limit does NOT tend to zero; so \((-1, 3) \notin S\).
• \((-1, 1)\):
  \[ 2 + (-1) \times 1 = 2 - 1 = 1 > 0 \] Limit tends to zero; so \((-1,1) \in S\).
• \((1, -1)\):
  \[ 2 + 1 \times (-1) = 2 -1 = 1 > 0 \] Limit tends to zero; so \((1,-1) \in S\).
• \((1, -2)\):
  \[ 2 + 1 \times (-2) = 2 - 2 = 0 \] Limit does NOT tend to zero (limit may be finite nonzero or undefined); so \((1,-2) \notin S\).

Final Answer:
\[ (-1,1) \in S \quad \text{and} \quad (1,-1) \in S \] Only these two points satisfy the limit condition.