Problem 18 - A Complete Resource Book in Mathematics for JEE Main

Question

The values of constants \(a\) and \(b\) so that $$ \lim _{x \rightarrow \infty}\left(\frac{x^{2}+1}{x+1}-a x-b\right)=0 \text { are } $$ (A) \(a=1, b=-1\) (B) \(a=-1, b=1\) (C) \(a=0, b=0\) (D) \(a=2, b=-1\)

Step-by-Step Solution

Answer

The values are (A) \(a=1, b=-1\).

Step 1: Simplify the Expression

Consider the expression within the limit: \( \frac{x^2+1}{x+1} - ax - b \). First, simplify \( \frac{x^2+1}{x+1} \) using polynomial long division. Divide \( x^2 + 1 \) by \( x + 1 \). The quotient is \( x - 1 \) with a remainder of 2. This means:\[\frac{x^2+1}{x+1} = x - 1 + \frac{2}{x+1}.\]Thus, the expression becomes \( x - 1 + \frac{2}{x+1} - ax - b \).

Step 2: Combine Like Terms

Rewrite the expression \( x - 1 + \frac{2}{x+1} - ax - b \) by combining like terms. Separate it into parts based on the powers of \( x \):\[(1-a)x + (-1-b) + \frac{2}{x+1}.\]

Step 3: Evaluate the Limit Condition

For \( \lim_{x \to \infty} ((1-a)x + (-1-b) + \frac{2}{x+1}) = 0 \) to be true, each component of the expression that grows as \( x \to \infty \) must cancel out. Hence, the coefficient of \( x \), \( (1-a) \), must be zero, and the constant term \((-1-b)\) must also be zero.

Step 4: Solve the System of Equations

From the limit condition, we derived two equations: \( 1-a = 0 \) and \( -1-b = 0 \). Solve these equations:\[\begin{align*}1-a &= 0 \Rightarrow a = 1, \-1-b &= 0 \Rightarrow b = -1.\end{align*}\]

Step 5: Conclude the Answer

With \( a = 1 \) and \( b = -1 \), the condition for the limit is satisfied. Therefore, the correct answer is (A) \( a = 1, b = -1 \).

Key Concepts

Polynomial Long Division

Polynomial long division is a method used to divide polynomials similar to arithmetic long division. It's an essential tool in calculus for simplifying expressions, especially when dealing with limits as variables approach infinity or other values.

Here's how it works:

• **Identify the Dividend and Divisor:** In polynomial division, the polynomial you want to divide is the dividend (e.g., \(x^2 + 1\)), and the term you are dividing by is the divisor (e.g., \(x + 1\)).
• **Divide the Leading Terms:** Divide the leading term of the dividend by the leading term of the divisor to find the first term of the quotient. For instance, dividing \(x^2\) by \(x\) gives \(x\).
• **Multiply and Subtract:** Multiply the entire divisor by the quotient obtained in the previous step and subtract it from the original dividend. This reduces your dividend to a smaller degree polynomial.
• **Repeat the Process:** Repeat the process using the new polynomial as the dividend until you obtain a remainder that has a lesser degree than the divisor.

In our exercise, dividing \(x^2 + 1\) by \(x + 1\) simplifies the expression to \(x - 1 + \frac{2}{x+1}\). This simplification allows us to evaluate limits more easily.
Understanding polynomial long division can simplify complex expressions and help in various calculus applications, making it a fundamental skill to master.

Asymptotic Behavior

In calculus, asymptotic behavior describes how a function behaves as it approaches a particular value or infinity. This is particularly useful when determining limits.

To identify asymptotic behavior, consider the terms of a function that have the most significant growth rate as the variable heads towards a limit. In our example, we have a rational expression \(\frac{x^2 + 1}{x + 1}\) where the numerator and denominator are polynomials.

As \(x\) approaches infinity:

• **Dominant Terms:** Only the highest degree terms in the numerator and denominator significantly affect the behavior of the function. Here, \(x^2\) in the numerator and \(x\) in the denominator dominate.
• **Simplification:** Often, we divide each term by the highest degree term in the denominator to simplify the expression, making it easier to evaluate the limit.
• **Behavior Analysis:** In this case, \(\frac{x^2}{x}\) simplifies to \(x\), and the asymptote's behavior points towards a linear expression influenced by this simplification.

Asymptotic behavior provides crucial insights, especially when combined with calculations like polynomial division, to solve limit problems effectively.

System of Equations

A system of equations consists of multiple equations that are solved together to find a common solution. In calculus, systems of equations frequently arise as a natural consequence of setting conditions for limits or derivatives.

In the problem at hand, solving for constants \(a\) and \(b\) involves:

• **Deriving Equations:** Based on the limit condition, we establish equations: \(1-a = 0\) and \(-1-b = 0\).
• **Simultaneous Solution:** These equations indicate that the expressions must independently satisfy the condition that they tend to zero at infinity.
  
  
• **Solve Sequentially:** Each equation represents a constraint that values of \(a\) and \(b\) must meet. Solving them gives directly \(a = 1\) and \(b = -1\).

Systems of equations are critical for translating complex calculus problems into simpler algebraic forms, allowing for a straightforward resolution of unknowns. This foundational skill aids in comprehensively understanding higher-level calculus concepts.