Question

Evaluate the limit \(\lim _{x \rightarrow a} \frac{x-a}{x^{3}-a^{3}}, \quad a \neq 0\).

Short answer

The limit is \(\frac{1}{3a^2}\).

Step-by-step solution

Substitute Directly

Start by substituting the value of \(x\) with \(a\) in the function \(\frac{x-a}{x^3-a^3}\). This gives us the indeterminate form \(\frac{0}{0}\), indicating that we need to simplify the expression.

Factor the Denominator

Recall that \(x^3 - a^3\) can be factored using the difference of cubes formula: \(x^3-a^3 = (x-a)(x^2+ax+a^2)\). Therefore, the function becomes \(\frac{x-a}{(x-a)(x^2+ax+a^2)}\).

Cancel Common Factors

Notice that \(x-a\) is a common factor in the numerator and denominator. Cancel this factor to simplify the function to \(\frac{1}{x^2+ax+a^2}\).

Evaluate the Limit

Now substitute \(x = a\) into the simplified expression \(\frac{1}{x^2+ax+a^2}\). This gives \(\frac{1}{a^2+a^2+a^2} = \frac{1}{3a^2}\). Thus, the limit is \(\frac{1}{3a^2}\).

Key concepts

Indeterminate Forms

When we first evaluate a limit by direct substitution, we might encounter expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These are known as indeterminate forms because they don't have a definite value. In calculus, indeterminate forms suggest that we need to perform additional algebraic manipulations to find the limit. In this exercise, substituting \( x = a \) directly resulted in \( \frac{0}{0} \), which means the expression is indeterminate, and further simplification is required. This opens the door to various techniques, such as factoring or rationalizing, to resolve these forms. By recognizing an indeterminate form, you know you're on the right path but need to do more work to get the answer.

Factoring Polynomials

Factorization is a key technique in algebra that breaks down a polynomial into simpler parts or factors. Factoring helps simplify expressions, especially when dealing with limits that might include indeterminate forms. In this case, we used the polynomial \( x^3 - a^3 \), which can be factored using a special algebraic identity. The factored form is \( (x-a)(x^2+ax+a^2) \). By factoring, we can often cancel out terms, which can simplify the expression significantly, making it easier to substitute and evaluate the limit. Remember that being familiar with factoring techniques is essential in solving calculus problems efficiently.

Difference of Cubes

The difference of cubes is a specific type of factorization used for expressions like \( x^3 - a^3 \). The formula for factoring the difference of cubes is:

• \( x^3 - a^3 = (x-a)(x^2 + ax + a^2) \)

This pattern of factoring is crucial when dealing with problems involving cubic expressions. Once we apply the difference of cubes, the polynomial is split into two parts: a binomial and a trinomial. The benefit of this is evident in this limit problem, where the binomial \( x-a \) appears in the numerator as well, allowing for cancellation. Recognizing and applying the difference of cubes formula will often simplify complex algebraic expressions.

Simplifying Algebraic Expressions

Simplifying algebraic expressions involves reducing them to their simplest form. This often means performing operations like factoring, canceling out common terms, and combining like terms. In our exercise, we identified \( x-a \) as a common factor in both the numerator and denominator. By canceling \( x-a \), the expression simplifies from \( \frac{x-a}{(x-a)(x^2+ax+a^2)} \) to \( \frac{1}{x^2+ax+a^2} \). Simplification is not just about making the math more manageable; it often makes it possible to evaluate limits that were previously indeterminate. Once simplified, direct substitution can yield a finite result, as it did here, leading to the solution \( \frac{1}{3a^2} \). Pay close attention to opportunities to simplify expressions, as they often pave the way to a clear and concise solution.