Let S be the set of all (α, β) ∈ R × R such that lim x→∞ sin(x^2)(logex)^α sin(1/x^2) / x^αβ(loge(1+x))^β = 0

Question

Let \(S\) be the set of all \((\alpha, \beta) \in \mathbb{R} \times \mathbb{R}\) such that 

\(\lim\limits _{x \rightarrow \infty} \frac{\sin \left(x^{2}\right)\left(\log _{e} x\right)^{\alpha} \sin \left(\frac{1}{x^{2}}\right)}{x^{\alpha \beta}\left(\log _{e}(1+x)\right)^{\beta}}=0\)

Then which of the following is (are) correct?

(A) \((-1,3) \in S\)

(B) \((-1,1) \in S\)

(C) \((1,-1) \in S\)

(D) \((1,-2) \in S\)

Answer 1

Correct options are (B) \((-1,1) \in S\) and (C) \((1,-1) \in S\)

\(\lim\limits _{x \rightarrow \infty} \frac{\sin \left(x^{2}\right) \sin \left(\frac{1}{x^{2}}\right)(\ln x)^{\alpha}}{x^{\alpha \beta}(\ln (1+x))^{\beta}}=0\)

\(=\lim\limits _{x \rightarrow \infty} \frac{\left(\sin x^{2}\right) \sin \left(\frac{1}{x^{2}}\right) \frac{1}{x^{2}}}{\left(\frac{1}{x^{2}}\right) x^{\alpha \beta}(\ln (1+x))^{\beta}}=0\)

It is possible if \(\alpha \beta+2>0\)

\(\alpha \beta+2>0\)

(A) \(\alpha \beta=-3\)

(B) \(\alpha \beta=-1\)

(C) \(\alpha \beta=-1\)

(D) \(\alpha \beta=-2\)