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

Step 1: Given Limit 

The given limit is: \[ \lim_{x \to \infty} \frac{\sin(x^2) (\log_e x)^\alpha \sin \left( \frac{1}{x^2} \right)} {x^{\alpha \beta} (\log_e (1 + x))^\beta}. \]

Step 2: Approximation for Large \( x \)

As \( x \to \infty \), using the small-angle approximation: \[ \sin \left( \frac{1}{x^2} \right) \approx \frac{1}{x^2}. \] Substituting this: \[ \lim_{x \to \infty} \frac{(\log_e x)^\alpha}{x^{\alpha \beta} (\log_e (1 + x))^\beta} \cdot \frac{1}{x^2}. \]

Step 3: Substituting \( \log_e x = t \)

Let \( \log_e x = t \), then: \[ x = e^t. \] This transforms our expression into: \[ \frac{t^\alpha}{e^{t \alpha \beta} t^\beta} \cdot e^{-2t}. \] Simplifying: \[ t^\alpha e^{-t(\alpha \beta + 2)}. \]

Step 4: Evaluating the Limit

For the limit to be **0**, we require: \[ \alpha \beta + 2 > 0 \quad \Rightarrow \quad \alpha \beta > -2. \]

Step 5: Checking Given Options

• For \( (-1, 3) \) : \( \alpha \beta = (-1)(3) = -3 \) (**Not Valid**).
• For \( (-1, 1) \) : \( \alpha \beta = (-1)(1) = -1 \) (**Valid**).
• For \( (1, -1) \) : \( \alpha \beta = (1)(-1) = -1 \) (**Valid**).
• For \( (1, -2) \) : \( \alpha \beta = (1)(-2) = -2 \) (**Not Valid**).

Final Conclusion

The values of \( (\alpha, \beta) \) that satisfy the condition \( \alpha \beta > -2 \) are: (-1,1) and (1,-1).