Problem 46 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The value of \(\lim _{n \rightarrow \infty} \frac{\sqrt[4]{n^{5}+2}-\sqrt[3]{n^{2}+1}}{\sqrt[5]{n^{4}+2}-\sqrt[2]{n^{3}+1}}\) is (A) \(I\) (B) 0 (C) \(-1\) (D) \(\infty\)

Step-by-Step Solution

Answer

The limit is 0.

Step 1: Analyze the limit form

We have a limit in the form \( \lim_{n \rightarrow \infty} \frac{f(n)}{g(n)} \), where \( f(n) = \sqrt[4]{n^5+2} - \sqrt[3]{n^2+1} \) and \( g(n) = \sqrt[5]{n^4+2} - \sqrt[2]{n^3+1} \). As \( n \) approaches infinity, each root expression tends towards infinity. Therefore, this is an indeterminate form \( \frac{\infty}{\infty} \). We need to simplify each part to find the limit.

Step 2: Simplify numerator \( f(n) \)

First, consider \( \sqrt[4]{n^5+2} \). This can be approximated by \( n^{5/4} \) because the term \( +2 \) becomes negligible as \( n \) grows. Similarly, \( \sqrt[3]{n^2+1} \approx n^{2/3} \) for large \( n \). Thus, \( f(n) \approx n^{5/4} - n^{2/3} \). Now compare the powers: \( \frac{5}{4} \approx 1.25 \) and \( \frac{2}{3} \approx 0.67 \), implying \( n^{5/4} \) will dominate for large \( n \).

Step 3: Simplify denominator \( g(n) \)

For the denominator, \( \sqrt[5]{n^4+2} \approx n^{4/5} \) and \( \sqrt[2]{n^3+1} \approx n^{3/2} \). For large \( n \), \( g(n) \approx n^{4/5} - n^{3/2} \). Comparing these, \( \frac{3}{2} = 1.5 \) and \( \frac{4}{5} = 0.8 \) imply that \( n^{3/2} \) will be the dominant term.

Step 4: Evaluate the limit

The dominant term for the numerator is \( n^{5/4} \), and for the denominator is \(-n^{3/2} \). Thus, the limit becomes:\[ \lim_{n \rightarrow \infty} \frac{n^{5/4}}{-n^{3/2}} = \lim_{n \rightarrow \infty} -n^{5/4 - 3/2} \]Since \( 5/4 - 3/2 = -0.25 \), this reduces to:\[ \lim_{n \rightarrow \infty} -n^{-0.25} = \lim_{n \rightarrow \infty} -\frac{1}{n^{0.25}} = 0 \]

Step 5: Conclude the limit value

Since \( n^{-0.25} \rightarrow 0 \) as \( n \rightarrow \infty \), the entire expression approaches 0. Thus, the limit is 0.

Key Concepts

Indeterminate Forms

Indeterminate forms are when the value of a limit becomes unclear or undefined given initial expressions. These often appear when attempting direct substitution in limits involving infinity.In this exercise, the limit initially takes on an indeterminate form of \(\frac{\infty}{\infty}\).This arises when both the numerator and the denominator grow indefinitely as the variable \(n\) approaches infinity.
When faced with such forms, it's crucial to simplify or transform the expression.This allows you to determine the limit accurately.By using techniques like algebraic manipulation or identifying dominant terms, you can resolve indeterminate limits effectively.

Algebraic Manipulation

Algebraic manipulation involves simplifying complex expressions to make limits more tractable.This can include approximating terms that significantly impact the limit when \(n\) becomes large.
For this exercise, we approximate:

• \( \sqrt[4]{n^5+2} \approx n^{5/4} \)
• \( \sqrt[3]{n^2+1} \approx n^{2/3} \)
• \( \sqrt[5]{n^4+2} \approx n^{4/5} \)
• \( \sqrt[2]{n^3+1} \approx n^{3/2} \)

These simplifications allow us to identify which terms have a major impact as the variable grows larger.This leads us away from the \(\frac{\infty}{\infty}\) form to something more generally workable.
By comparing the exponents, you can determine which terms will dominate the behavior of the function as \(n\) increases indefinitely.

Dominant Terms in Limits

When evaluating limits, identifying dominant terms helps to understand how expressions behave asymptotically.In simple terms, dominant terms significantly outshine others within an expression as the variable becomes very large.
In this exercise:

• The dominant term for \(f(n)\) is \(n^{5/4}\) because it has the highest power relative to \(n^{2/3}\).
• For \(g(n)\), \(n^{3/2}\) dominates over \(n^{4/5}\).

Knowing these dominant terms allows us to simplify the given limit's expression by focusing on these power terms.From here, the limit transforms from \( \frac{n^{5/4}}{-n^{3/2}} \) to \(-n^{-0.25} = -\frac{1}{n^{0.25}}\).
As \(n\) approaches infinity, \(-\frac{1}{n^{0.25}}\) approaches zero.This simplification reduces the problem into an easily solvable form, allowing us to conclude the limit's value accurately.