Question

Evaluate \(\lim _{x \rightarrow 1} \frac{\sqrt[3]{x}-1}{x-1} .\) (Hint: \(x-1=(\sqrt[3]{x})^{3}-1^{3}\) ).

Short answer

$$ Answer: The value of the limit is $$\frac{1}{3}.$$

Step-by-step solution

Identify the indeterminate form

We are given the expression $$\lim_{x\rightarrow 1}\frac{\sqrt[3]{x}-1}{x-1}.$$ First, let's check whether it is in indeterminate form by trying to evaluate the limit: $$\lim_{x\rightarrow 1}\frac{\sqrt[3]{x}-1}{x-1} = \frac{\sqrt[3]{1}-1}{1-1} = \frac{0}{0}.$$ Indeed, it's in the indeterminate form of \(\frac{0}{0}\).

Factor and cancel using the given hint

We are given the hint that \(x-1=(\sqrt[3]{x})^{3}-1^{3}\). Expanding it: $$x-1 = x^{1/3^3} - 1 = x^{1/3^3} - 1^{1/3^3},$$ we get the difference between two cubes, which can be factored using the formula for difference of cubes, giving us: $$x-1 = (x^{1/3}-1)\cdot(x^{2/3}+x^{1/3}+1).$$ Now, we'll rewrite the expression by substituting the given hint: $$\lim_{x\rightarrow 1}\frac{\sqrt[3]{x}-1}{(x^{1/3}-1)(x^{2/3}+x^{1/3}+1)}.$$ After substituting the hint, notice that we can now cancel the \((\sqrt[3]{x}-1)\) term in the numerator and the denominator: $$\lim_{x\rightarrow 1}\frac{1}{x^{2/3}+x^{1/3}+1}.$$

Evaluate the limit

Now, our function is no longer in indeterminate form and we can directly evaluate the limit as \(x\) approaches \(1\): $$\lim_{x\rightarrow 1}\frac{1}{x^{2/3}+x^{1/3}+1} = \frac{1}{1^{2/3}+1^{1/3}+1} = \frac{1}{1+1+1} = \frac{1}{3}.$$ After the steps mentioned above, we found that $$\lim_{x\rightarrow 1}\frac{\sqrt[3]{x}-1}{x-1} = \frac{1}{3}.$$

Key concepts

Indeterminate Form

When studying calculus, particularly the concept of limits, students often encounter expressions that lead to what's called an indeterminate form. This occurs when the limit of a function results in a form such as \( \frac{0}{0} \), which lacks a clear meaning in mathematical analysis. It's essential to recognize that though this form does not have an immediate value, it doesn't signify that the limit doesn't exist or is zero. Instead, indeterminate forms signal the need for further manipulation to reveal the true nature of the limit.

Indeterminate forms are not limited to \( \frac{0}{0} \) but include other types like \( \frac{\infty}{\infty} \), \( 0 \times \infty \), \( \infty - \infty \), \( 0^0 \), \( \infty^0 \), and \( 1^\infty \). To effectively analyze and simplify these forms, various strategies such as factoring, conjugation, or L'Hôpital's Rule can be employed.

Factoring in Calculus

Factoring is a powerful algebraic tool that comes in handy in calculus, especially when dealing with challenging limits like indeterminate forms. Factoring refers to the process of breaking down a complex expression into simpler, more manageable terms (factors) that, when multiplied together, reconstruct the original expression. In the context of limits, factoring can provide clarity by canceling out common terms in the numerator and the denominator of a rational function.

In the example shown in the textbook solution, factoring is crucial to resolving the indeterminate form. When the limit as \(x\) approaches 1 generates \( \frac{0}{0} \) in the original function, we're nudged to examine the structure of the denominator. Recognizing patterns such as the difference of cubes allows transformation of the limit, enabling simplification and cancellation of like terms. The outcome reveals the behavior of the function around the point that was initially ambiguous, thus providing a concrete answer to the limit.

Difference of Cubes

The difference of cubes is an algebraic factorization method particularly helpful when dealing with polynomial limits in calculus. This specific pattern arises when you have a difference between two terms, each raised to the third power: \( a^3 - b^3 \). The factorization of this expression follows the formula \( a^3 - b^3 = (a-b)(a^2 + ab + b^2) \).

Knowing this formula is advantageous because it transforms an unwieldy cubic difference into a product of a linear and a quadratic term, which oftentimes simplifies the evaluation of limits. In the exercise, the expression \( x - 1\) is rewritten using this principle, recognizing that \( x \) and \( 1 \) are the cubes of \( \sqrt[3]{x} \) and \( 1 \), respectively. Applying the difference of cubes formula results in a factored form that readily cancels out, clearing the pathway toward finding the limit. These algebraic manipulations, while they may seem routine, are vital for successfully interpreting and solving more complex calculus problems involving limits.