Problem 439 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

$$ \text { Find the constants } \left.a \text { and } b \text { such that } \lim _{x \rightarrow \infty} \frac{x^{2}+1}{x+1}-a x-b=0 \quad \text { Ans. } a=1, b=-1\right\\} $$

Step-by-Step Solution

Answer

The constants \(a\) and \(b\) that satisfy the given limit condition are \(a=1\) and \(b=-1\).

Step 1: Rewrite the given limit

Rewrite the given expression so that the subtraction is inside the limit, this better allows us to see how the limit will act on the equation. So, \[ \lim_{x \rightarrow \infty} \frac{x^{2}+1}{x+1}-a x-b=0 \] becomes \[ \lim_{x \rightarrow \infty} ( \frac{x^{2}+1}{x+1}-a x-b)=0 \]

Step 2: Simplify the equation

To simplify this, we can perform polynomial division or use the fact that as \(x \rightarrow \infty\), \(x^{2}/x\) is much larger than 1/x. Thus, we can relatively ignore the 1 and rewrite the equation as: \[ \lim_{x \rightarrow \infty} (x-a x+b)=0 \]

Step 3: Compare Coefficients

To solve for a and b, equate the coefficients of like terms in the equation. From \(x - ax + b = 0\), it's clear that the coefficient of \(x\) must be 0, implying that \(a = 1\). We can also see that the constant term must be 0 and therefore, \(b = -1\).

Key Concepts

Understanding Limits at Infinity

Figuring out what happens to mathematical expressions as numbers get larger and larger is a recurring theme in calculus, and this is where the concept of finding limits at infinity comes into play. It deals with the behavior of a function as the input value—or variable—approaches infinity. When we look at the expression \( \lim\text{\textunderscore}{x \rightarrow \infty} \frac{x^{2}+1}{x+1}-ax-b=0 \) from the original exercise, we're basically trying to understand what values of \(a\) and \(b\) would make the expression tend towards zero as \(x\) becomes larger and larger.

This kind of problem can represent, for example, the behavior of a physical system as time goes to infinity, or the value of a financial investment as the number of years increases indefinitely. In the context of our original limit problem, determining the constants \(a\) and \(b\) ensures that the mathematical expression will remain balanced—or equal to zero—as \(x\) grows without bound.

Polynomial Long Division

To simplify complex rational expressions, especially where the degrees of polynomials differ significantly, we often turn to polynomial long division. It's akin to the long division of numbers we all learn in grade school, but instead of dealing with numbers, it manipulates polynomials. This method allows us to divide the numerator by the denominator to either simplify the expression or find the remainder.

In our exercise, one approach to finding the limit would be to use polynomial long division to divide \(x^2 + 1\) by \(x + 1\) continuously until you no longer have a remainder, simplifying the equation. The result of this division would then allow us to find the right coefficients, \(a\) and \(b\), that satisfy the equation at infinity. While our step-by-step solution ultimately opted for a faster approximation method due to the behavior of the function as \(x\) increases, knowing how to perform polynomial long division remains a fundamental and versatile tool in calculus.

Coefficient Comparison

When solving problems in calculus, particularly those involving limits or polynomials, the technique of coefficient comparison becomes invaluable. This involves equating the coefficients of corresponding powers of \(x\) on both sides of an equation to solve for unknown constants. In essence, for two polynomials to be equal, the coefficients for each exponent must match.

In our exercise, after the simplification of the limit expression, we arrived at a stage where we had \(x - ax + b\). Because we know that this expression must equal zero no matter the value of \(x\), we compared coefficients to determine that the coefficient of \(x\) must be zero, thus \(a-1=0\), which leads us to conclude that \(a = 1\). Similarly, since there is no \(x^0\) or constant term on the right side of the equation, to balance the equation the coefficient \(b\) must also lead to zero, giving us \(b = -1\). Coefficient comparison is a straightforward yet powerful strategy to untangle the complexities of algebraic expressions in calculus.