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

Short answer

Answer: The limit is \(11 + 4\sqrt{2}\).

Step-by-step solution

Recognize the given expression as a difference of two powers

To start, notice that the given limit expression is in the form of a difference of two powers, i.e., \(x^{\frac{1}{4}}-2^{\frac{1}{4}}=x^{\frac{1}{4}} - 2^{\frac{1}{4}}\).

Apply the factorization formula

Using the factorization formula provided, \(x^n - a^n = (x-a)(x^{n-1}+x^{n-2}a+...+xa^{n-2}+a^{n-1})\), and taking \(n=\frac{1}{4}\) and \(a=2\), we have: $$x^{\frac{1}{4}} - 2^{\frac{1}{4}} = \left(x-2^{\frac{1}{4}}\right)\left(x^{\frac{1}{4}-1}+x^{\frac{1}{4}-2}2^{\frac{1}{4}}+x^{\frac{1}{4}-3}(\sqrt{2})+2^{\frac{1}{4}-1}\right)$$

Simplify the expression

Simplifying the above expression, $$x^{\frac{1}{4}} - 2^{\frac{1}{4}} = \left(x-2^{\frac{1}{4}}\right)\left(x^{\frac{3}{4}}+x^{\frac{1}{2}}2^{\frac{1}{4}}+x^{\frac{1}{4}}(\sqrt{2})+2^{\frac{1}{2}}\right)$$

Replace the given expression in the limit

Now, we replace the expression in the limit with the simplified form: $$\lim _{x \rightarrow 16} \frac{\sqrt[4]{x}-2}{x-16} = \lim _{x \rightarrow 16} \frac{\left(x-2^{\frac{1}{4}}\right)\left(x^{\frac{3}{4}}+x^{\frac{1}{2}}2^{\frac{1}{4}}+x^{\frac{1}{4}}(\sqrt{2})+2^{\frac{1}{2}}\right)}{(x-16)}$$

Factor out (x-16)

Factor out \((x-16)\) from the expression: $$\lim _{x \rightarrow 16} \frac{\left(x-16\right)\left(x^{\frac{3}{4}-1}+x^{\frac{1}{2}-1}2^{\frac{3}{4}}+x^{\frac{1}{4}-1}(\sqrt[4]{16})+2^{\frac{1}{2}-1}\right)}{(x-16)}$$

Cancel out (x-16)

Cancel out \((x-16)\) in the numerator and denominator to get: $$\lim _{x \rightarrow 16} \left(x^{\frac{1}{4}}+x^{\frac{1}{2}}2^{\frac{1}{4}}+x^{\frac{1}{4}}(\sqrt{2})+2^{\frac{1}{2}}\right)$$

Evaluate the limit

As \(x \rightarrow 16\), $$\lim _{x \rightarrow 16} \left(x^{\frac{1}{4}}+x^{\frac{1}{2}}2^{\frac{1}{4}}+x^{\frac{1}{4}}(\sqrt{2})+2^{\frac{1}{2}}\right) = 16^{\frac{1}{4}}+16^{\frac{1}{2}}2^{\frac{1}{4}}+16^{\frac{1}{4}}(\sqrt{2})+2^{\frac{1}{2}} = 2+8+4\sqrt{2}+1$$ Thus, the limit is equal to \(11 + 4\sqrt{2}\)

Key concepts

Factorization Formula

The factorization formula is a vital tool when dealing with expressions resembling the difference of powers, especially in calculus when solving limits. The formula is structured as follows:

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

This equation facilitates breaking down complex expressions into simpler components, enabling easier manipulation when computing limits.
In our exercise, you are asked to compute a limit involving the expression \(\sqrt[4]{x} - 2\). Notice that this expression can be seen as a difference of powers with a fractional exponent. Applying the formula essentially involves recognizing the expression's structure to simplify it before evaluating. The power of the factorization formula lies in its ability to transform a convoluted polynomial difference into terms that are manageable and ready for further operations, such as cancellations in limits.

Difference of Powers

Understanding the term 'difference of powers' is crucial when approaching seemingly complex expressions in calculus.
A 'difference of powers' expression typically follows the form \(x^{n} - a^{n}\), where both terms are raised to an exponent. In limits, such expressions often appear due to their tendency to initially result in an indeterminate form when directly substituting the limit value.

• For example, consider the limit expression \(\lim_{x \to 16} \frac{\sqrt[4]{x} - 2}{x-16}\).
• It features a difference of powers where the exponents are fractional, arising from the fourth root of \(x\) against a constant.

The strategy to overcome this form of indeterminacy involves either factorization (as discussed) or transforming the expression through other techniques to allow for direct evaluation.

Rationalizing Techniques

Rationalizing, although traditionally associated with removing radicals from denominators, is a broader technique applicable in limit problems. It can be likened to factorization in that it seeks to restructure an expression into a more evaluable form.

• In problems featuring roots or fractional powers like \(\sqrt[4]{x}\), rationalizing techniques might involve multiplying the numerator and denominator by a conjugate or related term to clear radicals or enable simplification.
• This often aids in allowing terms to cancel out, thus resolving indeterminate forms.

In the exercise, while direct rationalization isn't explicitly performed, the concept of transforming expressions to eliminate complexities remains integral. It's about making the problem simpler and revealing the underlying mathematical structure that allows for easy handling of the limit as \(x\) approaches a particular value.