Problem 76 - Calculus Early Transcendentals

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. $$\lim _{x \rightarrow 16} \frac{\sqrt[4]{x}-2}{x-16}$$

Step-by-Step Solution

Answer

Answer: No, the limit does not exist as x approaches 16.

Step 1: Rewrite the expression in terms of \((x^n - a^n)\)

To apply the factorization formula, we need to rewrite the given expression as \((x^n - a^n)\). In this case, we have $$\frac{\sqrt[4]{x}-2}{x-16}=\frac{x^{\frac{1}{4}}-2^{\frac{1}{4}}}{x-2^4}$$ Now we can see that this is a case of the given formula with \(n=\frac{1}{4}\) and \(a=2\).

Step 2: Apply the factorization formula to simplify the expression

Now we apply the factorization formula: $$x^{\frac{1}{4}}-2^{\frac{1}{4}}=(x-2)\left(x^{\frac{1}{4}-1}+x^{\frac{1}{4}-\frac{2}{4}}\cdot2+x^{\frac{1}{4}-\frac{3}{4}}\cdot 2^2+\cdots+x^{\frac{1}{4}-\frac{4}{4}}\cdot 2^{3}\right)$$ The given expression simplifies as: $$\frac{(x-2)\left(x^{\frac{1}{4}-1}+x^{\frac{1}{4}-\frac{2}{4}}\cdot2+x^{\frac{1}{4}-\frac{3}{4}}\cdot 2^2+\cdots+x^{\frac{1}{4}-\frac{4}{4}}\cdot 2^{3}\right)}{x-2^4}$$ We notice that there is a common factor of \((x-2)\) in the numerator and denominator, so we can cancel it out: $$\frac{(x^{\frac{1}{4}-1}+x^{\frac{1}{4}-\frac{2}{4}}\cdot2+x^{\frac{1}{4}-\frac{3}{4}}\cdot 2^2+\cdots+x^{\frac{1}{4}-\frac{4}{4}}\cdot 2^{3})}{x-16}$$

Step 3: Calculate the limit as x approaches 16

Now we can evaluate the limit as x approaches 16: $$\lim_{x \rightarrow 16} \frac{x^{\frac{1}{4}-1}+x^{\frac{1}{4}-\frac{2}{4}}\cdot2+x^{\frac{1}{4}-\frac{3}{4}}\cdot 2^2+\cdots+x^{\frac{1}{4}-\frac{4}{4}}\cdot 2^{3}}{x-16}$$ $$=\frac{16^{\frac{1}{4}-1}+16^{\frac{1}{4}-\frac{2}{4}}\cdot 2+16^{\frac{1}{4}-\frac{3}{4}}\cdot 2^2+\cdots+16^{\frac{1}{4}-\frac{4}{4}}\cdot 2^{3}}{16-16}$$ Evaluating the terms in the expression: $$=\frac{4^{-1}+2+2^3}{16-16}$$ $$=\frac{1/4+2+8}{0}$$ which is not defined. Therefore, the limit does not exist.

Key Concepts

Factorization Formula

The factorization formula is a powerful tool in calculus, especially when dealing with polynomial expressions. The essence of the formula is to break down a higher power polynomial into a simpler form, which aids in simplifying otherwise complicated expressions. In the exercise, we use this formula to tackle limits involving polynomials, where variables are powered and need simplification.
The formula states that for any given positive integer \( n \) and a real number \( a \), the expression \( x^n - a^n \) can be factorized as:

• \((x-a)(x^{n-1} + x^{n-2}a + x^{n-3}a^2 + \,\cdots\, + xa^{n-2} + a^{n-1})\)

Applying this concept to solve limits involves recognizing patterns and rewriting the expressions accordingly. Here, we rewrite the polynomial in terms of this formula so we can effectively cancel out terms and simplify the problem.

Limit Calculation

Limit calculation is a core concept in calculus. It allows us to understand the behavior of functions as they approach a certain point, often providing crucial information even when the function itself is not defined at that point. Calculating the limit typically involves direct substitution, algebraic manipulation, or using a series of proven techniques.
In our specific problem, the limit calculation begins by transforming the given expression using the factorization formula. After rewriting it and simplifying the expression, we can then directly find what value the function approaches as \( x \) nears 16.
However, it's important to notice that calculating limits sometimes leads to indeterminate forms like \( \frac{0}{0} \). It's crucial to handle these carefully, as they often signal a misstep in the simplification process or demand further manipulation to resolve.

Algebraic Manipulation

Algebraic manipulation is the art of rearranging and simplifying algebraic expressions to make calculations more manageable. It is an essential skill in solving limits and involves factoring, expanding, combining like terms, and canceling out factors when possible.
To effectively solve a problem involving limits like in this exercise, algebraic manipulation is used to rewrite the original expression in a form that makes it easier to solve. This often involves:

• Factoring expressions to expose and cancel common terms
• Simplifying complex fractions
• Using known techniques like the factorization formula to breakdown complicated expressions

Through careful and strategic algebraic manipulation, seemingly complex problems become solvable, allowing for the calculation of limits where direct methods might fail.