Problem 92 - Calculus: Early Transcendentals

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

Step-by-Step Solution

Answer

Question: Calculate the limit of the given rational function as x approaches -1, using the factorization formula for \(x^7 - a^7\) with \(a=-1\). Answer: Given the rational function, the limit as x approaches -1 is 5.

Step 1: Write down the given limit

We are given the limit: $$\lim _{x \rightarrow-1} \frac{x^{7}+1}{x+1}$$

Step 2: Rewrite the expression using given hint

Using the hint provided, we have to use the factorization formula for \(x^7 - a^7\) where \(a = -1\). So, the expression becomes: $$\lim_{x \rightarrow -1}\frac{x^7 - (-1)^7}{x+1} = \lim_{x \rightarrow -1}\frac{x^7 - (-1)}{x+1}$$

Step 3: Apply the factorization formula

Recall the given factorization formula: $$x^{n}-a^{n}=(x-a)\left(x^{n-1}+a x^{n-2}+a^{2} x^{n-3}+\cdots+a^{n-2} x+a^{n-1}\right)$$ Applying this formula for \(x^7 - (-1)\): $$x^7+1 = (x+1)\left(x^6-x^5+x^4-x^3+x^2-x+1\right)$$ Now we can rewrite the limit using this factorization: $$\lim_{x \rightarrow -1}\frac{x^7+1}{x+1} = \lim_{x \rightarrow -1} \frac{(x+1)\left(x^6-x^5+x^4-x^3+x^2-x+1\right)}{x+1}$$

Step 4: Simplify the expression and evaluate the limit

Now that we have a factor of \((x+1)\) in both the numerator and denominator, we can cancel them out. This leaves us with: $$\lim_{x \rightarrow -1} \left(x^6-x^5+x^4-x^3+x^2-x+1\right)$$ Now, we can directly substitute \(x = -1\) into this expression: $$(-1)^6-(-1)^5+(-1)^4-(-1)^3+(-1)^2-(-1)+1 = 1-1+1-1+1+1+1 = 5$$ Thus, our final answer is: $$\lim _{x \rightarrow-1} \frac{x^{7}+1}{x+1}=5$$

Key Concepts

Factorization Formula

The factorization formula is a powerful tool in limit calculations, especially when dealing with polynomial expressions. It allows us to simplify expressions by factoring, which often helps in removing indeterminate forms like \(\frac{0}{0}\). The specific formula used here is for the difference of powers, given by:- \(x^{n} - a^{n} = (x-a)(x^{n-1} + ax^{n-2} + a^{2}x^{n-3} + \cdots + a^{n-2}x + a^{n-1})\)This formula is particularly useful when \(x\) approaches a value where the polynomial expression becomes zero. The root \(x = a\) can be extracted using the formula, helping to cancel terms in a fraction and effectively simplifying the calculation. In our problem, using the hint provided, the exercise required substituting \(a = -1\) into the formula for \(x^7 - a^7\), transforming the expression into a product that simplifies the limit.

Polynomial Division

Polynomial division is a technique similar to long division but involves polynomials instead of numbers. In limit calculations, it's a strategy used to simplify expressions by dividing polynomials, essentially breaking down high-degree polynomials into more manageable forms. When using the factorization formula, polynomial division often plays an indirect role as it simplifies the expression after factoring.For instance, in the example given, the expression \(\frac{x^7 + 1}{x+1}\) was simplified by canceling out \(x+1\), which is a result of implicit polynomial division after factorization. This simplification is crucial because it leaves us with a polynomial that no longer yields an indeterminate form at \(x = -1\). This reduced form is much simpler to evaluate.

Substitution Method

The substitution method in calculus involves replacing a variable with a specific value to directly calculate the limit. Particularly if the division method results in a simplified form, substituting directly into the reduced polynomial makes finding the limit far easier.After factoring and simplifying the expression in this exercise, the step of direct substitution was essential. By substituting \(x = -1\) into the polynomial \((x^6-x^5+x^4-x^3+x^2-x+1)\), avoiding the original indeterminate form was possible, allowing for straightforward evaluation:- Replace \(x\) with \(-1\) in the polynomial: \((-1)^6 - (-1)^5 + (-1)^4 - (-1)^3 + (-1)^2 - (-1) + 1\)- Compute each power: every odd power of \(-1\) is \(-1\), and every even power is \(1\).- Sum these results: \(1 - 1 + 1 - 1 + 1 + 1 + 1 = 5\).Thus, substitution allows us to evaluate the simplified form directly, efficiently leading to the final limit value.