Problem 6 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\(\lim _{n \rightarrow \infty} \frac{n^{k} \sin ^{2}(n !)}{n+2} 0

Step-by-Step Solution

Answer

The limit is 0 (Option C).

Step 1: Analyze the Expression

The given limit expression is \( \lim _{n \rightarrow \infty} \frac{n^{k} \sin^{2}(n!)}{n+2} \). We need to determine the behavior as \( n \rightarrow \infty \), especially focusing on \( n^k \) and \( \sin^2(n!) \).

Step 2: Split the Terms

Consider the term \( \frac{n^k}{n+2} \) separately from \( \sin^2(n!) \), since \( \sin^2(n!) \) is bounded between 0 and 1. Hence, the expression becomes: \( \lim _{n \rightarrow \infty} \frac{n^k}{n+2} \sin^2(n!) \).

Step 3: Simplify \( \frac{n^k}{n+2} \)

As \( n \rightarrow \infty \), the expression \( \frac{n^k}{n+2} \) approaches \( n^{k-1} \) because \( n+2 \approx n \). Since \( 0 < k < 1 \), \( k-1 < 0 \), making \( n^{k-1} \rightarrow 0 \).

Step 4: Consider the Overall Limit

Since \( \frac{n^k}{n+2} \rightarrow 0 \) and \( \sin^2(n!) \) remains bounded, the product \( \frac{n^k \sin^2(n!)}{n+2} \rightarrow 0 \cdot [0, 1] = 0 \).

Step 5: Conclude the Limit

The overall limit is determined by the dominance of \( \frac{n^k}{n+2} \rightarrow 0 \), leading to \( \lim _{n \rightarrow \infty} \frac{n^k \sin^2(n!)}{n+2} = 0 \).

Key Concepts

Asymptotic Behavior

Understanding the asymptotic behavior of a sequence involves examining how a sequence behaves as it approaches infinity. In the given problem, we need to evaluate how the expression \( \frac{n^k \sin^2(n!)}{n+2} \) behaves as \( n \) becomes very large. This involves breaking down the expression into parts that can be easily analyzed separately. The term \( n^k \) grows as \( n \) increases, but because \( k < 1 \), its growth is slower than linear. Meanwhile, the \( n+2 \) in the denominator grows linearly. Therefore, \( \frac{n^k}{n+2} \) simplifies to \( n^{k-1} \), which tends towards zero since \( k-1 < 0 \). This simplification helps us understand the overarching behavior of the expression, because multiplying this by a bounded \( \sin^2(n!) \) provides clarity on how it diminishes as \( n \) becomes infinitely large.

Trigonometric Functions

Trigonometric functions, like \( \sin(\theta) \), have known properties that help simplify complex limits. One important property is that \( \sin(\theta) \) always lies between -1 and 1, meaning \( \sin^2(\theta) \) will range between 0 and 1. For our sequence, \( \sin^2(n!) \) fluctuates within this range irrespective of \( n \). This bounded nature significantly influences the limit of the sequence product \( \frac{n^k \sin^2(n!)}{n+2} \). When coupled with a part of the expression that tends toward zero, the overall product is dominated by the diminishing factor \( \frac{n^k}{n+2} \), ensuring that the entire sequence tends towards zero as \( n \to \infty \). This property of trigonometric functions ensures that unpredictable fluctuations do not affect the asymptotic result of the limit.

Exponent Properties

Exploring exponent properties is key to unraveling the behavior of the expression \( n^k \) within the problem. Since \( k \) is a fraction between 0 and 1, \( n^k \) grows slower than \( n \). This property indicates that \( n^{k-1} \), or \( \frac{n^k}{n} \), effectively tends to zero. This is because the subtraction \( k-1 \) results in a negative exponent, changing the term into \( \frac{1}{n^{1-k}} \). As \( n \) grows, \( \frac{1}{n^{1-k}} \) approaches zero, overpowering any minor influence from any other bounded component. By dissecting the exponent's properties, we recognize why \( \frac{n^k}{n+2} \) simplifies to zero, and solidify the impact it has on the sequence towards reaching the limit conclusion of zero.