Problem 26 - Calculus 1 Course in Mathematics for the IIT JEE and Other Engineering Exams

Question

6\. \(\operatorname{Lt}_{x \rightarrow 1} \frac{x^{n}+x^{n-1}+x^{n-2}+\ldots \ldots+x^{2}+x-n}{x-1}\) is equal to (a) \(\frac{n(n+1)}{2}\) (b) \(\frac{n(n-1)}{2}\) (c) \(\frac{n(n+2)}{2}\) (d) None of these

Step-by-Step Solution

Answer

The limit evaluates to \( \frac{n(n+1)}{2} \). The correct option is (a).

Step 1: Recognize the form

The limit \( \operatorname{Lt}_{x \rightarrow 1} \frac{x^{n}+x^{n-1}+x^{n-2}+\ldots+x^{2}+x-n}{x-1} \) resembles the limit form of \( \frac{0}{0} \) because when \( x = 1 \), both numerator and denominator become zero.

Step 2: Identify the function for the numerator

The numerator \( x^{n} + x^{n-1} + x^{n-2} + \ldots + x^{2} + x - n \) is a polynomial in variable \( x \) with the subtraction of constant \( n \). When evaluated specifically at \( x = 1 \), it simplifies to \( n - n = 0 \).

Step 3: Apply L'Hôpital's Rule if applicable

Given it's a \( \frac{0}{0} \) form, we should use L'Hôpital's Rule. Differentiate the numerator and the denominator with respect to \( x \). The derivative of the denominator, \( x - 1 \), is 1. Now, differentiate the numerator.

Step 4: Differentiate the numerator

The derivative of the polynomial \( x^{n} + x^{n-1} + x^{n-2} + \ldots + x^{2} + x \) with respect to \( x \) results in \( nx^{n-1} + (n-1)x^{n-2} + (n-2)x^{n-3} + \ldots + 2x + 1 \). The derivative of \( n \) is zero as it is a constant.

Step 5: Evaluate the derivative at \( x = 1 \)

Substitute \( x = 1 \) into the derivative obtained. This results in the expression \( n \cdot 1^{n-1} + (n-1)\cdot 1^{n-2} + \ldots + 2 \cdot 1 + 1 \). Each term becomes 1, and the series becomes \( n + (n-1) + \ldots + 2 + 1 \).

Step 6: Simplify the arithmetic series

The series \( n + (n-1) + \ldots + 2 + 1 \) is an arithmetic series. The sum of this series is \( \frac{n(n+1)}{2} \).

Step 7: Choose the correct answer

Among the given options, option (a) \( \frac{n(n+1)}{2} \) matches the derived limit result.

Key Concepts

L'Hôpital's Rule

In calculus, L'Hôpital's Rule is a method for finding limits that result in indeterminate forms such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). The basic idea is to simplify these forms by differentiating the numerator and the denominator separately.
It's crucial to realize that L'Hôpital's Rule can only be applied to limits of quotients where both the top and bottom approach zero, or both approach infinity.

• Step 1: Verify the form \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
• Step 2: Differentiate the numerator and the denominator.
• Step 3: Take the limit of the new quotient.

This technique simplifies the limit of complicated functions, making it easier to handle expressions encountered in calculus problems.

Polynomial Differentiation

Polynomial differentiation involves finding the derivative of a polynomial function. Polynomials are expressions composed of variables and coefficients, arranged in powers combined using addition, subtraction, and multiplication.
The process of differentiation allows us to calculate the slope of a polynomial and is done using simple rules:

• The derivative of \( x^n \) with respect to \( x \) is \( nx^{n-1} \).
• Constants have a derivative of zero.
• Derivatives of sums are the sums of derivatives.

In the problem, differentiating the polynomial \( x^{n} + x^{n-1} + ... + x - n \) gives us \( nx^{n-1} + (n-1)x^{n-2} + ... + 1 \). This forms the basis for using L'Hôpital's Rule effectively.

Indeterminate Forms (0/0)

Indeterminate forms, by their nature, create uncertainty in mathematical calculations. The \( \frac{0}{0} \) form, for example, occurs when both the numerator and denominator of a fraction approach zero as \( x \) approaches some limit.
This concept is pivotal in calculus because it requires additional tools, such as L'Hôpital's Rule, for resolution. Without knowing what this form approaches, predictions or conclusions aren't possible using basic rules of fractions. Recognizing an indeterminate form helps us know when more advanced techniques are needed.

Sum of Arithmetic Series

An arithmetic series is the sum of terms in a sequence where each term after the first is a constant increment more than the previous term. For instance, in the series \( n + (n-1) + ... + 2 + 1 \), the increment is -1.
The sum \( S \) of the first \( n \) numbers in an arithmetic series can be calculated using:\[ S = \frac{n}{2} (a + l) \]where \( a \) is the first term, and \( l \) is the last term. For the series derived in our problem, the formula changes slightly to \( \frac{n(n+1)}{2} \), neatly solving our limit's arithmetic portion and simplifying our understanding of the problem's context.