Let $S$ be the set of all $(\alpha, \beta) \in \mathbb{R} \times \mathbb{R}$ such that $$\lim_{x \rightarrow \infty} \frac{\sin \left(x^{2}\right)\left(\log _{e} x\right)^{\alpha} \sin \left(\frac{1}{x^{2}}\right)}{x^{\alpha \beta}\left(\log _{e}(1+x)\right)^{\beta}}=0 .$$ Then which of the following is (are) correct ?

(B) $(-1,1) \in S$; (C) $(1,-1) \in S$

Limits — JEE Advanced · 2024 · 26 May 2024 · Paper 2 Question with Solution

Let

S

be the set of all

(α, β) \in R \times R

such that

x \to \infty lim \frac{sin ( x ^{2} ) ( lo g _{e} x ) ^{α} sin ( \frac{1}{x ^{2}} )}{x ^{α β} ( lo g _{e} ( 1 + x ) ) ^{β}} = 0.

Then which of the following is (are) correct ?

Answer: (B), (C)

Using the approximations

sin (x^{2}) \approx 1

(bounded),

sin (\frac{1}{x ^{2}}) \approx \frac{1}{x ^{2}}

x \to \infty

, and

lo g_{e} (1 + x) \approx lo g_{e} x

x \to \infty lim \frac{( lo g _{e} x ) ^{α}}{x ^{α β + 2} ( lo g _{e} x ) ^{β}} = x \to \infty lim \frac{( lo g _{e} x ) ^{α - β}}{x ^{α β + 2}} = 0

Substituting

t = lo g_{e} x

t \to \infty lim \frac{t ^{α - β}}{e ^{t (α β + 2)}} = 0

This limit is zero when

α β + 2 > 0

, i.e.,

α β > - 2

.

Checking the options:

Therefore, (B) and (C) are correct.

JEE Advanced · 2024 · 26 May 2024 · Paper 2 · One or more correct options