Problem 33 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The value of \(\lim _{x \rightarrow 1} \frac{x^{n}+x^{n-1}+x^{n-2}+\ldots+x^{2}+x-n}{x-1}\) is (A) \(\frac{n(n+1)}{2}\) (B) 0 (C) 1 (D) \(n\)

Step-by-Step Solution

Answer

The limit evaluates to \(\frac{n(n+1)}{2}\), which is option (A).

Step 1: Identify the Limit Expression

The given limit expression is \( \lim _{x \rightarrow 1} \frac{x^{n} + x^{n-1} + x^{n-2} + \ldots + x^{2} + x - n}{x-1} \). This is a polynomial divided by \(x-1\).

Step 2: Recognize the Pattern

The numerator represents the sum of a geometric series \(x^n + x^{n-1} + \ldots + x + 1\) evaluated at \(x=1\), minus \(1\). The expression can be rewritten as \(\lim_{x \to 1} \frac{(x^{n+1} - 1)/(x-1) - 1}{x-1}\).

Step 3: Use L'Hopital's Rule

Since the direct substitution of \(x = 1\) in the expression results in an indeterminate form \(\frac{0}{0}\), apply L'Hopital's Rule. Differentiate the numerator and the denominator separately with respect to \(x\).

Step 4: Differentiate the Numerator and Denominator

The derivative of the numerator, after simplification, is the sum \((nx^{n-1} + (n-1)x^{n-2} + \ldots + 2x + 1) - 0\). The derivative of the denominator \(x-1\) is \(1\).

Step 5: Evaluate the Derivative at x=1

After differentiating and simplifying, substitute \(x = 1\) into the derived expression to evaluate it. The expression becomes \(n + (n-1) + \ldots + 2 + 1 = \frac{n(n+1)}{2}\).

Step 6: Select the Correct Answer

The evaluated limit at \(x = 1\) coincides with option A, which is \(\frac{n(n+1)}{2}\).

Key Concepts

Geometric Series

A geometric series is a sequence of numbers where each term after the first is found by multiplying the previous one by a fixed, non-zero number called the common ratio. In a finite geometric series like the one in the given limit expression, we have terms extending from \(x^n\) down to \(x^0\).
To understand the concept better, the formula for the sum of a finite geometric series can be utilized, which is:

• \(S = a \frac{1-r^n}{1-r}\)

Here \(a\) is the first term and \(r\) is the common ratio.
In the context of our problem, the geometric series itself doesn't change; it is evaluated at \(x=1\). When \(x\) approaches 1, the series simplifies significantly because the sum of all coefficients minus 1 leads to indirect involvement of zeros causing \(\frac{0}{0}\) condition. This indeterminate form is handled using calculus tools.

L'Hopital's Rule

L'Hôpital's Rule is an essential calculus tool that helps find limits involving indeterminate forms like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). When directly substituting the point of interest in a limit results in these forms, L'Hôpital's Rule suggests that we can differentiate the numerator and the denominator separately and then try to evaluate the limit again.
In our specific exercise, applying \(x=1\) resulted in \(\frac{0}{0}\). By differentiating the numerator \((x^n + x^{n-1} + \ldots + x + 1) - n\) with respect to \(x\), we get each term reduced by one power and thus leading to a computation of a simple polynomial sum.
After this simplification, calculating the limit becomes straightforward, allowing us to directly evaluate at \(x=1\) and easily find the answer \(\frac{n(n+1)}{2}\).

Polynomial Division

Polynomial division is the process of dividing a polynomial by another polynomial of lower or equal degree, often resulting in a quotient and a remainder. It is analogous to long division with numbers.
In the case of the limit \(\lim _{x \rightarrow 1} \frac{x^{n}+x^{n-1}+x^{n-2}+\ldots+x^{2}+x-n}{x-1}\), the numerator can be seen as a large polynomial polynomial with respect to \(x-1\). The act of considering the limit as \(x\) approaches 1 can be related to testing the remainder behavior when such polynomials are divided. This is closely related to the factor theorem, where any polynomial \(f(x)\) such that \(x-c\) is a factor will have \(f(c) = 0\), leading us to derive the expression as a series divided by \(x-1\).
With the application of the derivative in L'Hopital's Rule, we skillfully bypass the traditional polynomial division here, simplifying our evaluative work considerably.