Question

Useful factorization formula 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. $$\lim _{x \rightarrow-1} \frac{x^{7}+1}{x+1}$$ (Hint: Use the formula for \(\left.x^{7}-a^{7} \text { with } a=-1 .\right)\)

Short answer

Answer: The limit of the expression as x approaches -1 is 7.

Step-by-step solution

Since we are given \(x^7 - a^7\), let's change it to \(x^7 + 1\) by substituting \(a = -1\). In this case, \(x^7 + 1 = x^7 - (-1)^7\) and \(n = 7\). Now, update the factorization formula: $$x^7 - (-1)^7 = (x - (-1))(x^6 + x^5(-1) + x^4(-1)^2 + x^3(-1)^3 + x^2(-1)^4 + x(-1)^5 + (-1)^6)$$ #Step 2: Simplify the formula# Now simplify the formula by plugging in the values of \(a\) and \(n\).

After substituting the values of \(a\) and \(n\), we get: $$x^7 + 1 = (x + 1)(x^6 - x^5 + x^4 - x^3 + x^2 - x + 1)$$ #Step 3: Find the limit using the simplified equation# Next, we will use the above simplified equation to find the limit of the given expression and simplify.

Our goal is to find the limit of the following expression: $$\lim_{x\rightarrow -1} \frac{x^7 + 1}{x + 1}$$ Using the simplified equation, we can rewrite the expression as: $$\lim_{x\rightarrow -1} \frac{(x + 1)(x^6 - x^5 + x^4 - x^3 + x^2 - x + 1)}{x + 1}$$ Now, we can cancel out the \((x + 1)\) terms in the numerator and denominator: $$\lim_{x\rightarrow -1} (x^6 - x^5 + x^4 - x^3 + x^2 - x + 1)$$ Finally, substitute the value of \(x\) approaching -1 into the equation: $$((-1)^6 - (-1)^5 + (-1)^4 - (-1)^3 + (-1)^2 - (-1) + 1)$$ #Step 4: Simplify the expression to find the limit# Lastly, simplify the expression to find the limit.

After substituting the value of \(x\) and evaluating, we obtain: $$(1 + 1 + 1 + 1 + 1 + 1 + 1)$$ The final answer is: $$\lim_{x\rightarrow -1} \frac{x^7 + 1}{x + 1} = 7$$

Key concepts

Limits

Understanding limits is a fundamental concept in calculus. A limit describes how a function behaves as the input, or variable, approaches a specific value. In our case, we deal with the limit as \( x \) approaches -1.
When solving limits involving polynomials, it's often about simplifying the expression until you can directly substitute the value of \( x \).
For instance, in the given problem, the mathematical expression \( \lim_{x \rightarrow -1} \frac{x^7 + 1}{x + 1} \) initially seems complex due to division by zero when directly substituting \( x = -1 \).

• By factoring out, we can cancel out terms and simplify the limit expression.
• Once simplified, substituting \( x = -1 \) becomes possible without causing undefined terms or zero-division errors.

Appreciating this process is vital, as it reflects many real-world applications where predicting behaviors as conditions change is crucial.

Polynomials

Polynomials are expressions formed by variables raised to whole-number exponents, combined using addition, subtraction, and multiplication. They are versatile and appear frequently in calculus problems.
In the original exercise, we see an example of a polynomial: \( x^7 + 1 \). This can be related to an important polynomial identity, which helps factorize expressions smoothly.

• The identity \( x^n - a^n = (x-a)(x^{n-1} + x^{n-2}a + \ldots + a^{n-1}) \) allows for such transformation and simplification.
• It transforms cumbersome polynomial division into a manageable multiplication, which can be systematically simplified.

This characteristic of polynomials makes them a core component of calculus, where simplifying complex expressions often relies on polynomial manipulation skills.

Calculus

Calculus is the branch of mathematics focused on change and motion, utilizing foundational principles like limits, derivatives, and integrals.
The problem given, though seemingly simple with polynomial division, encapsulates important calculus concepts such as limits.

• It demonstrates the technique of transforming seemingly indeterminate forms into determinate simplified expressions.
• The exercise exemplifies how calculus starts with simple algebraic manipulations before venturing into deeper analyses like rate of changes and area under curves.

While the problem solves a polynomial division by using a factorization method, understanding it within the broader context of calculus methodology is crucial for deeper math studies and applications.