Let $S$ be the set of all $(\alpha, \beta) \in \mathbb{R} \times \mathbb{R}$ such that $\lim_{x \to \infty} \frac{\sin(x^2) (\log_e x)^{\sin \left( \frac{1}{x^2} \right)}}{x^{\alpha\beta} (\log_e (1+x))^{\beta}} = 0$. Then which of the following is(are) correct? Options: (A) $(-1, 3) \in S$ (B) $(-1, 1) \in S$ (C) $(1, -1) \in S$ (D) $(1, -2) \in S$

### Analyzing the Setup When you see expressions like $$\lim_{x \to \infty} \frac{\sin(x^2) (\ln x)^{\sin(1/x^2)}}{x^{\alpha\beta} (\ln(1+x))^\beta} = 0$$ your first instinct might be to reach for L'Hopital's Rule. Stop! That is a trap. In JEE Advanced, we value elegance over brute force. Let us use the power of asymptotic analysis. ### Taming the Numerator First, look at $\sin(1/x^2)$. As $x \to \infty$, the argument $1/x^2$ shrinks to zero. We know that for small $\theta$, $\sin \theta \approx \theta$. So, we can safely replace $\sin(1/x^2)$ with $1/x^2$. Now, consider the term $(\ln x)^{\sin(1/x^2)}$. Let us call it $L$. Taking the natural log, we get: $$\ln L = \sin(1/x^2) \cdot \ln(\ln x)$$ Substituting our approximation, this becomes: $$\ln L \approx \frac{\ln(\ln x)}{x^2}$$ Here is the secret: polynomial growth ($x^2$) is the king of the jungle. It grows much faster than any logarithmic function. Thus, $\frac{\ln(\ln x)}{x^2} \to 0$ as $x \to \infty$. Since $\ln L \to 0$, our term $L$ must approach $e^0 = 1$. The numerator is now effectively $1$. ### The Denominator's Secret Now for the denominator: $\ln(1+x)$. We can rewrite this as $\ln(x(1 + 1/x)) = \ln x + \ln(1 + 1/x)$. As $x \to \infty$, the term $\ln(1 + 1/x)$ approaches $\ln(1) = 0$. So, $\ln(1+x)$ behaves exactly like $\ln x$. Our entire limit expression has now collapsed into a beautiful, simple form: $$\lim_{x \to \infty} \frac{1}{x^{\alpha\beta} (\ln x)^\beta}$$ ### The Golden Condition Let us clean that up: $$\lim_{x \to \infty} \frac{1}{x^{\alpha\beta} (\ln x)^\beta} = 0$$ For this fraction to equal zero, the denominator must grow to infinity. The term $(\ln x)^\beta$ is slow, but the term $x^{\alpha\beta}$ is powerful. If the exponent $\alpha\beta$ is positive, the denominator explodes to infinity, and the limit becomes zero. If it is negative, the $x$ term moves to the numerator, and the limit becomes infinity. If it is zero, we are left with a power of $\ln x$, which also goes to infinity. Therefore, the only condition for the limit to be zero is **$\alpha\beta > 0$**. ### Conclusion By testing our options, we find that values satisfying **$\alpha\beta > 0$** are the correct solutions. You have just mastered the art of asymptotic dominance. Keep this intuition, and no limit will ever scare you again!

JEE(ADVANCED)-202

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $S$ be the set of all $(α, β) \in R \times R$ such that $lim_{x \to \infty} \frac{s i n ( x ^{2} ) ( l o g _{e} x ) ^{s i n (\frac{1}{x ^{2}})}}{x ^{α β} ( l o g _{e} ( 1 + x ) ) ^{β}} = 0$ . Then which of the following is(are) correct?

Select Answer:

* Multiple Correct

Progressive Hints

Hint 1

Hint 1: The Spark. When you see a limit involving

x \to \infty

with complex functions like

sin (1/ x^{2})

and logarithms, do not panic. Your first step is to isolate the behavior of each component. Ask yourself: as

x

becomes massive, what does

sin (1/ x^{2})

look like? Use the standard limit

lim_{θ \to 0} \frac{s i n θ}{θ} = 1

to simplify the trigonometric part.

Hint 2

Hint 2: The Arsenal. You need to handle the term

(ln x)^{s i n (1/ x^{2})}

. This is an indeterminate form of the type

1^{0}

or similar. Take the natural logarithm of this expression, say

L

, to get

ln L = sin (1/ x^{2}) \cdot ln (ln x)

. Now, analyze the growth rates. Does the polynomial

x^{2}

in the denominator of the sine approximation overpower the logarithmic term

ln (ln x)

Hint 3

Hint 3: The Execution. After simplifying all terms, you should arrive at a limit of the form

lim_{x \to \infty} \frac{1}{x ^{α β + 2} ( l n x ) ^{β}} = 0

. For this fraction to approach zero, the denominator must grow to infinity. This happens only if the exponent of

x

is strictly positive. Set up the inequality

α β + 2 > 0

and test your options.

Visualized Solution

$lim_{x \to \infty} f (x) = 0$

Given Limit: $lim_{x \to \infty} \frac{s i n ( \frac{1}{x ^{2}} ) ( l o g _{e} x ) ^{s i n (\frac{1}{x ^{2}})}}{x ^{α β} ( l o g _{e} ( 1 + x ) ) ^{β}} = 0$
Goal: Find valid pairs of $(α, β)$ from the given options.
Method: Analyze the asymptotic behavior of each term as $x \to \infty$ .

$sin (\frac{1}{x ^{2}})$ as $x \to \infty$

Consider the first term: $sin (\frac{1}{x ^{2}})$
As $x \to \infty$ , the argument $\frac{1}{x ^{2}} \to 0$ .
Using standard limit $lim_{θ \to 0} \frac{s i n θ}{θ} = 1$ , we get $sin θ \sim θ$ .
Therefore, $sin (\frac{1}{x ^{2}}) \sim \frac{1}{x ^{2}}$ .

$(lo g_{e} x)^{s i n (1/ x^{2})}$ as $x \to \infty$

Consider the second term: $L = (lo g_{e} x)^{s i n (\frac{1}{x ^{2}})}$
Take natural logarithm: $ln L = sin (\frac{1}{x ^{2}}) ln (lo g_{e} x)$
Using previous approximation: $ln L \sim \frac{l n ( l o g _{e} x )}{x ^{2}}$
As $x \to \infty$ , polynomial growth outpaces logarithmic growth, so $\frac{l n ( l o g _{e} x )}{x ^{2}} \to 0$ .
Thus, $ln L \to 0 ⟹ L \to e^{0} = 1$ .

$lo g_{e} (1 + x)$ as $x \to \infty$

Consider the denominator term: $lo g_{e} (1 + x)$
Factor out $x$ : $lo g_{e} (x (1 + \frac{1}{x}))$
Use log properties: $lo g_{e} x + lo g_{e} (1 + \frac{1}{x})$
As $x \to \infty$ , $\frac{1}{x} \to 0$ , so $lo g_{e} (1 + \frac{1}{x}) \to lo g_{e} 1 = 0$ .
Therefore, $lo g_{e} (1 + x) \sim lo g_{e} x$ .

Substituting Approximations

Original Limit: $lim_{x \to \infty} \frac{s i n ( \frac{1}{x ^{2}} ) ( l o g _{e} x ) ^{s i n (\frac{1}{x ^{2}})}}{x ^{α β} ( l o g _{e} ( 1 + x ) ) ^{β}}$
Substitute numerator approximations: $\frac{1}{x ^{2}} \cdot 1$
Substitute denominator approximations: $x^{α β} (lo g_{e} x)^{β}$
The limit becomes: $lim_{x \to \infty} \frac{x ^{- 2}}{x ^{α β} ( l o g _{e} x ) ^{β}}$

Simplifying the Expression

Combine the powers of $x$ in the denominator.
Move $x^{- 2}$ to the denominator: $x^{α β} \cdot x^{2} = x^{α β + 2}$
Simplified Limit: $lim_{x \to \infty} \frac{1}{x ^{α β + 2} ( l o g _{e} x ) ^{β}} = 0$

Condition for Zero Limit

We have $lim_{x \to \infty} \frac{1}{x ^{α β + 2} ( l o g _{e} x ) ^{β}} = 0$ .
For a fraction with a constant numerator to approach $0$ , its denominator must approach $\infty$ .
Therefore, we require: $lim_{x \to \infty} x^{α β + 2} (lo g_{e} x)^{β} = \infty$ .

Analyzing the Dominant Term

We need $x^{α β + 2} (lo g_{e} x)^{β} \to \infty$ as $x \to \infty$ .
The polynomial term $x^{α β + 2}$ strictly dominates the logarithmic term $(lo g_{e} x)^{β}$ for any real $β$ .
For the product to approach $\infty$ , the power of $x$ must be strictly positive.
Condition: $α β + 2 > 0$ .

Evaluating Option A

Option A: $(α, β) = (- 1, 3)$
Calculate $α β$ : $(- 1) (3) = - 3$
Check condition $α β + 2 > 0$ :
$- 3 + 2 = - 1$
Since $- 1$ is not greater than $0$ , Option A is incorrect.

Evaluating Option B

Option B: $(α, β) = (- 1, 1)$
Calculate $α β$ : $(- 1) (1) = - 1$
Check condition $α β + 2 > 0$ :
$- 1 + 2 = 1$
Since $1 > 0$ , the condition is satisfied. Option B is correct.

Evaluating Option C

Option C: $(α, β) = (1, - 1)$
Calculate $α β$ : $(1) (- 1) = - 1$
Check condition $α β + 2 > 0$ :
$- 1 + 2 = 1$
Since $1 > 0$ , the condition is satisfied. Option C is also correct.

Evaluating Option D

Option D: $(α, β) = (1, - 2)$
Calculate $α β$ : $(1) (- 2) = - 2$
Check condition $α β + 2 > 0$ :
$- 2 + 2 = 0$
If $α β + 2 = 0$ , denominator is $(lo g_{e} x)^{- 2}$ .
As $x \to \infty$ , $(lo g_{e} x)^{- 2} \to 0$ , so the fraction $\to \infty$ . Option D is incorrect.

Key Takeaways

Correct Options: (B) $(- 1, 1)$ and (C) $(1, - 1)$ .
Key Takeaway 1: Always analyze the asymptotic behavior of individual terms in complex limits at infinity.
Key Takeaway 2: Polynomial growth ( $x^{k}, k > 0$ ) always dominates logarithmic growth ( $(lo g_{e} x)^{n}$ ).

00:00 / 00:00

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

When you see expressions like

x \to \infty lim \frac{sin ( x ^{2} ) ( ln x ) ^{s i n (1/ x^{2})}}{x ^{α β} ( ln ( 1 + x ) ) ^{β}} = 0

your first instinct might be to reach for L'Hopital's Rule. Stop! That is a trap. In JEE Advanced, we value elegance over brute force. Let us use the power of asymptotic analysis.

Taming the Numerator

First, look at

sin (1/ x^{2})

. As

x \to \infty

, the argument

1/ x^{2}

shrinks to zero. We know that for small

θ

sin θ \approx θ

. So, we can safely replace

sin (1/ x^{2})

with

1/ x^{2}

Now, consider the term

(ln x)^{s i n (1/ x^{2})}

. Let us call it

L

. Taking the natural log, we get:

ln L = sin (1/ x^{2}) \cdot ln (ln x)

Substituting our approximation, this becomes:

ln L \approx \frac{ln ( ln x )}{x ^{2}}

Here is the secret: polynomial growth (

x^{2}

) is the king of the jungle. It grows much faster than any logarithmic function. Thus,

\frac{l n ( l n x )}{x ^{2}} \to 0

x \to \infty

. Since

ln L \to 0

, our term

L

must approach

e^{0} = 1

. The numerator is now effectively

1

The Denominator's Secret

Now for the denominator:

ln (1 + x)

. We can rewrite this as

ln (x (1 + 1/ x)) = ln x + ln (1 + 1/ x)

x \to \infty

, the term

ln (1 + 1/ x)

approaches

ln (1) = 0

. So,

ln (1 + x)

behaves exactly like

ln x

Our entire limit expression has now collapsed into a beautiful, simple form:

x \to \infty lim \frac{1}{x ^{α β} ( ln x ) ^{β}}

The Golden Condition

Let us clean that up:

x \to \infty lim \frac{1}{x ^{α β} ( ln x ) ^{β}} = 0

For this fraction to equal zero, the denominator must grow to infinity. The term

(ln x)^{β}

is slow, but the term

x^{α β}

is powerful.

If the exponent

α β

is positive, the denominator explodes to infinity, and the limit becomes zero. If it is negative, the

x

term moves to the numerator, and the limit becomes infinity. If it is zero, we are left with a power of

ln x

, which also goes to infinity.

Therefore, the only condition for the limit to be zero is $α β > 0$ .

Conclusion

By testing our options, we find that values satisfying $α β > 0$ are the correct solutions. You have just mastered the art of asymptotic dominance. Keep this intuition, and no limit will ever scare you again!

Common Traps & FAQs

Wait, why can we just replace

sin (1/ x^{2})

with

1/ x^{2}

? Is that always allowed?

That is a fantastic question! We are using the asymptotic equivalence

sin θ \sim θ

θ \to 0

. Since

1/ x^{2} \to 0

x \to \infty

, the approximation is perfectly valid. In JEE Advanced, this is a powerful tool to simplify complex limits. Just ensure the argument of the sine function is indeed approaching zero!

I am confused about

(ln x)^{s i n (1/ x^{2})}

. Why does it approach 1?

Think of it this way: we took the natural log, which gave us

\frac{l n ( l n x )}{x ^{2}}

. As

x \to \infty

, the polynomial

x^{2}

grows infinitely faster than the iterated logarithm

ln (ln x)

. Because the denominator dominates, the fraction goes to 0. Since

ln L \to 0

, it follows that

L \to e^{0}

, which is 1. It is all about the hierarchy of growth rates!

Why does the condition

α β + 2 > 0

guarantee the limit is 0?

Great observation. We have a constant numerator (1) and a denominator that is a product of a power of

x

and a power of

ln x

. If the power of

x

is positive, the denominator blows up to infinity, making the whole fraction vanish to zero. If the power were negative, the

x

term would move to the numerator, making the limit infinity. Thus,

α β + 2 > 0

is our golden ticket.

Select Answer:

Visualized Solution

$lim_{x \to \infty} f (x) = 0$

$sin (\frac{1}{x ^{2}})$ as $x \to \infty$

$(lo g_{e} x)^{s i n (1/ x^{2})}$ as $x \to \infty$

$lo g_{e} (1 + x)$ as $x \to \infty$

Substituting Approximations

Simplifying the Expression

Condition for Zero Limit

Analyzing the Dominant Term

Evaluating Option A

Evaluating Option B

Evaluating Option C

Evaluating Option D

Key Takeaways

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

Taming the Numerator

The Denominator's Secret

The Golden Condition

Conclusion

Similar Questions

If $lim_{x \to 0} \frac{α e ^{x} + β e ^{- x} + γ s i n x}{x s i n ^{2} x} = \frac{2}{3}$ , where $α, β, γ \in R$ , then which of the following is NOT correct ?

$lim_{x \to 0} \frac{e ^{4∣ s i n x ∣} - 2∣ s i n x ∣ - 1}{x ^{2}}$

For $α, β, γ, \in R$ , if $lim_{x \to 0} \frac{x ^{2} s i n α x + ( γ - 1 ) e ^{x^{2}}}{s i n 2 x - β x} = 3$ , then $β + γ - α$ is equal to:

For each $t \in R$ , let [t] be the greatest integer less than or equal to t. Then, $lim_{x \to 1^{+}} \frac{( 1 - ∣ x ∣ + s i n ∣1 - x ∣ ) s i n ( \frac{π}{2} [ 1 - x ])}{∣1 - x ∣ [ 1 - x ]}$

If $lim_{x \to 0} \frac{α x e ^{x} - β l o g _{e} ( 1 + x ) + γ x ^{2} e ^{- x}}{x s i n ^{2} x} = 10$ , $α, β, γ \in R$ , then the value of $α + β + γ$ is

If $lim_{x \to \infty} (1 + \frac{a}{x} + \frac{b}{x ^{2}})^{2 x} = e^{2}$ , then the values of $a$ and $b$ , are

If $lim_{x \to 0} \frac{3 + α s i n x + β c o s x + l o g _{e} ( 1 - x )}{3 t a n ^{2} x} = \frac{1}{3}$ , then $2 α - β$ is equal to :

Let $α, β \in R$ be such that $lim_{x \to 0} \frac{x ^{2} s i n ( β x )}{α x - s i n x} = 1$ . Then $6 (α + β)$ equals .........

$lim_{x \to 0} \frac{x ( e ^{1 + x^{2} + x^{4} - 1} - 1 )}{1 + x ^{2} + x ^{4} - 1}$

The value of $lim_{x \to 0} ((sin x)^{1/ x} + (1/ x)^{s i n x})$ , where $x > 0$ is

Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limx→∞​f(x)=0

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin(x21​) as .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x→∞

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } (loge​x)sin(1/x2) as .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x→∞

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Substituting Approximations

Simplifying the Expression

Condition for Zero Limit

Analyzing the Dominant Term

Evaluating Option A

Evaluating Option B

Evaluating Option C

Evaluating Option D

Key Takeaways

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Analyzing the Setup

Taming the Numerator

The Denominator's Secret

The Golden Condition

Conclusion

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If limx→0​xsin2xαex+βe−x+γsinx​=32​, where α,β,γ∈R, then which of the following is NOT correct ?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limx→0​x2e4∣sinx∣−2∣sinx∣−1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For α,β,γ,∈R, if limx→0​sin2x−βxx2sinαx+(γ−1)ex2​=3, then β+γ−α is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For each t∈R, let [t] be the greatest integer less than or equal to t. Then, limx→1+​∣1−x∣[1−x](1−∣x∣+sin∣1−x∣)sin(2π​[1−x])​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If limx→0​xsin2xαxex−βloge​(1+x)+γx2e−x​=10, α,β,γ∈R, then the value of α+β+γ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If limx→∞​(1+xa​+x2b​)2x=e2, then the values of a and b, are

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If limx→0​3tan2x3+αsinx+βcosx+loge​(1−x)​=31​, then 2α−β is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let α,β∈R be such that limx→0​αx−sinxx2sin(βx)​=1. Then 6(α+β) equals .........

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limx→0​1+x2+x4​−1x(e1+x2+x4​−1−1)​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of limx→0​((sinx)1/x+(1/x)sinx), where x>0 is

$lim_{x \to \infty} f (x) = 0$

$sin (\frac{1}{x ^{2}})$ as $x \to \infty$

$(lo g_{e} x)^{s i n (1/ x^{2})}$ as $x \to \infty$

$lo g_{e} (1 + x)$ as $x \to \infty$

If $lim_{x \to 0} \frac{α e ^{x} + β e ^{- x} + γ s i n x}{x s i n ^{2} x} = \frac{2}{3}$ , where $α, β, γ \in R$ , then which of the following is NOT correct ?

$lim_{x \to 0} \frac{e ^{4∣ s i n x ∣} - 2∣ s i n x ∣ - 1}{x ^{2}}$

For $α, β, γ, \in R$ , if $lim_{x \to 0} \frac{x ^{2} s i n α x + ( γ - 1 ) e ^{x^{2}}}{s i n 2 x - β x} = 3$ , then $β + γ - α$ is equal to:

For each $t \in R$ , let [t] be the greatest integer less than or equal to t. Then, $lim_{x \to 1^{+}} \frac{( 1 - ∣ x ∣ + s i n ∣1 - x ∣ ) s i n ( \frac{π}{2} [ 1 - x ])}{∣1 - x ∣ [ 1 - x ]}$

If $lim_{x \to 0} \frac{α x e ^{x} - β l o g _{e} ( 1 + x ) + γ x ^{2} e ^{- x}}{x s i n ^{2} x} = 10$ , $α, β, γ \in R$ , then the value of $α + β + γ$ is

If $lim_{x \to \infty} (1 + \frac{a}{x} + \frac{b}{x ^{2}})^{2 x} = e^{2}$ , then the values of $a$ and $b$ , are

If $lim_{x \to 0} \frac{3 + α s i n x + β c o s x + l o g _{e} ( 1 - x )}{3 t a n ^{2} x} = \frac{1}{3}$ , then $2 α - β$ is equal to :

Let $α, β \in R$ be such that $lim_{x \to 0} \frac{x ^{2} s i n ( β x )}{α x - s i n x} = 1$ . Then $6 (α + β)$ equals .........

$lim_{x \to 0} \frac{x ( e ^{1 + x^{2} + x^{4} - 1} - 1 )}{1 + x ^{2} + x ^{4} - 1}$

The value of $lim_{x \to 0} ((sin x)^{1/ x} + (1/ x)^{s i n x})$ , where $x > 0$ is