Question

Calculate the following limits using the factorization formula \( x^{n}-a^{n}=(x-a)\left(x^{n-1}+x^{n-2} a+x^{n-3} a^{2}+\cdots+x a^{n-2}+a^{n-1}\right) \) where \(n\) is a positive integer and a is a real number. $$ \lim _{x \rightarrow-1} \frac{x^{7}+1}{x+1} \text { (Hint: Use the formula for } x^{7}-a^{7} \text { with } a=-1 . \text { ) } $$

Short answer

Question: Find the limit using the factorization formula provided: $$\lim_{x\rightarrow -1} \frac{x^7 + 1}{x + 1}$$ Answer: The limit of the given function as x approaches -1 is $$\lim_{x \rightarrow -1} \frac{x^7 + 1}{x + 1} = 5.$$

Step-by-step solution

Rewrite the function using the factorization formula

The function we need to find the limit of is $$ \frac{x^7 + 1}{x + 1}. $$ We can rewrite this expression using the factorization formula for \(x^n - a^n\), with \(n = 7\) and \(a = -1\). The factorization formula for this case is $$ x^7 - (-1)^7 = (x - (-1))(x^6 + x^5 (-1) + x^4 (-1)^2 + \cdots + x (-1)^6 + (-1)^7). $$ Simplify the expression: $$ x^7 + 1 = (x + 1)(x^6 - x^5 + x^4 - x^3 + x^2 - x + 1). $$ Now, rewrite the given function using this factorization: $$ \frac{x^7 + 1}{x + 1} = \frac{(x + 1)(x^6 - x^5 + x^4 - x^3 + x^2 - x + 1)}{x + 1}. $$

Simplify the function

Now that we have written the function using the factorization formula, we can simplify it by canceling the common factor \((x + 1)\): $$ \frac{(x + 1)(x^6 - x^5 + x^4 - x^3 + x^2 - x + 1)}{x + 1} = x^6 - x^5 + x^4 - x^3 + x^2 - x + 1. $$

Calculate the limit

Finally, we can find the limit of the simplified function as x approaches -1: $$ \lim_{x\rightarrow -1} (x^6 - x^5 + x^4 - x^3 + x^2 - x + 1) = (-1)^6 - (-1)^5 + (-1)^4 - (-1)^3 + (-1)^2 - (-1) + 1 = 1 + 1 + 1 - 1 + 1 + 1 + 1 = 5. $$ Thus, the limit of the given function as x approaches -1 is $$ \lim_{x \rightarrow -1} \frac{x^7 + 1}{x + 1} = 5.

Key concepts

Limit calculation

Understanding limits is a fundamental concept in calculus. They are used to define continuity, derivatives, and integrals, among others. A limit describes the value that a function approaches as the input approaches some value. In this exercise, we are calculating the limit of a polynomial expression as the variable approaches -1. This helps us understand how the function behaves near that specific point.
When you have a fraction like \( \frac{x^7 + 1}{x + 1} \), directly substituting \(-1\) in for \(x\) results in a \(\frac{0}{0}\) form, which is indeterminate. To resolve this, we manipulate the expression (using techniques like factorization) so it can be simplified to a form that avoids this issue.
After simplifying, the expression becomes a polynomial \( x^6 - x^5 + x^4 - x^3 + x^2 - x + 1 \). Now, substituting \(-1\) directly into this is straightforward, leading to an easy calculation. Thus, the limit is calculated as 5.

Polynomial factorization

Polynomial factorization is a key algebraic technique used to simplify expressions, making limits easier to calculate. In this case, we used the formula for the difference of powers: \[ x^n - a^n = (x-a)(x^{n-1} + x^{n-2}a + \ldots + xa^{n-2} + a^{n-1}) \]This formula allows us to rewrite polynomial expressions like \( x^7 + 1 \) in a form that reveals common factors.
Here, \( a = -1 \), which makes the expression \( x^7 - (-1)^7 = x^7 + 1 \). The factor of \( (x + 1) \) simplifies the rest of the expression. This kind of manipulation is essential because it helps to cancel out terms and prevent indeterminate forms in limit calculations.
Factorization moves complex polynomials into more manageable parts, which streamlines solving and calculating polynomials in calculus.

Algebraic manipulation

Algebraic manipulation is the process of rearranging and simplifying expressions using various algebraic rules and properties. In this exercise, algebraic manipulation comes into play after applying the factorization formula to rewrite \( x^7 + 1 \) as \( (x + 1)(x^6 - x^5 + x^4 - x^3 + x^2 - x + 1) \).
The algebraic strategy used here is to factor out common terms from the numerator and the denominator. By factoring \( (x + 1) \) out of the equation, and then canceling it, we avoid an indeterminate form (like \(\frac{0}{0}\)) and simplify the process of calculating the limit.
This simplification reduces the original complex expression into its elementary components, making further operations, like taking limits, straightforward. Mastering algebraic manipulation is crucial for handling calculus problems effectively, especially when they involve fractions and limits.