Question

Calculate the following limits using the factorization formula $$ x^{n}-a^{n}=(x-a)\left(x^{n-1}+a x^{n-2}+a^{2} x^{n-3}+\cdots+a^{n-2} x+a^{n-1}\right) $$ where n is a positive integer and a is a real number. $$ \lim _{x \rightarrow 1} \frac{x^{6}-1}{x-1} $$

Short answer

Answer: The limit of the given expression as \(x \rightarrow 1\) is \(6\).

Step-by-step solution

Apply the factorization formula

Using the provided factorization formula for \(x^n - a^n\), where \(n=6\) and \(a=1\), we can rewrite the expression inside the limit as: $$ x^{6}-1^{6}=(x-1)\left(x^{5}+1x^4+1^2x^3+\cdots+1^4x+1^5\right) $$

Simplify the expression

The expression in parentheses simplifies to: $$ x^{5}+x^4+x^3+x^2+x+1 $$ So the limit can be rewritten as: $$ \lim_{x \rightarrow 1} \frac{(x-1)(x^5+x^4+x^3+x^2+x+1)}{x-1} $$

Cancel the common term

The common term \((x-1)\) in the numerator and denominator can be cancelled. Thus, the limit simplifies to: $$ \lim_{x \rightarrow 1} (x^5+x^4+x^3+x^2+x+1) $$

Evaluate the limit

Now, by substituting the limit value \(x=1\) into the simplified expression, we get: $$ 1^5+1^4+1^3+1^2+1+1 = 1+1+1+1+1+1=6 $$ So, the limit of the given expression as \(x \rightarrow 1\) is \(6\).

Key concepts

Factorization Formula

In calculus, the factorization formula is a powerful tool, especially when working with limits involving polynomial expressions. For any polynomial raised to a power, say \(x^n - a^n\), we can use the factorization formula to break down the expressions into factors that are much easier to manage.

This is extremely useful in cases where direct substitution into the limit would cause an indeterminate form like \(0/0\). Factorization creates an opportunity to simplify the expression and potentially eliminate terms that cause undefined behavior. In the context of the exercise:
\[x^{n}-a^{n}=(x-a)(x^{n-1}+ax^{n-2}+a^{2}x^{n-3}+\cdots+a^{n-2}x+a^{n-1})\]
The expression on the right is an algebraic identity that showcases how any difference of nth powers can be turned into a product of two factors: \((x-a)\) and a sum involving the terms of \(x\) and \(a\) raised to different powers. Applying this to the textbook exercise makes evaluating the limit straightforward.

Evaluating Limits

Evaluating limits is a fundamental concept in calculus, wherein we seek to determine the value that a function approaches as the input approaches a particular point. Sometimes, a direct substitution of the value into a function can provide the limit. However, this is not always possible, especially when the direct substitution results in an indeterminate form such as \(0/0\), as often encountered with rational expressions.

Several techniques, such as factorization, l'Hôpital's rule, and algebraic manipulation, can be used to evaluate such limits. In the given exercise, factorization plays a key role. Once we factor the polynomial using the factorization formula, we often find terms that cancel out, allowing for a direct substitution without reaching an indeterminate form, which is precisely what we did to find that the limit as \(x\) approaches 1 is 6.

Understanding how to maneuver around these indeterminate forms is essential for correctly evaluating limits and grasping the behavior of functions near points of interest.

Polynomial Functions

Polynomial functions are expressions that consist of variables and coefficients, structured in a way that each term involves the variable raised to a non-negative integer power. A polynomial function can take the form:\[P(x) = a_nx^n + a_{n-1}x^{n-1} + \ldots + a_2x^2 + a_1x + a_0\]
Where \(a_n, a_{n-1}, \ldots, a_0\) are constants with \(a_n eq 0\), and \(n\) is a non-negative integer known as the degree of the polynomial.

In our exercise, we deal with a polynomial function where the limit is sought as \(x\) approaches a specific value. Polynomials are continuous functions, which means they don't have any breaks or holes. This continuity makes evaluating limits for polynomials usually straightforward, as they often result in the polynomial being evaluated at that specific value, provided that no indeterminate forms are encountered. The simplicity of polynomials is apparent in the cancellation step of the solved exercise.