Problem 79 - University Calculus: Early Transcendentals

Question

Find a value of \(c\) that makes the function $$f(x)=\left\\{\begin{array}{ll} \frac{9 x-3 \sin 3 x}{5 x^{3}}, & x \neq 0 \\\c, & x=0\end{array}\right.$$. continuous at \(x=0 .\) Explain why your value of \(c\) works.

Step-by-Step Solution

Answer

The function is continuous at \(x = 0\) if \(c = -\frac{27}{10}\).

Step 1: Define Continuity at a Point

A function is continuous at a point, say \(x = a\), if the limit of the function as \(x\) approaches \(a\) from both sides is equal to the function value at \(a\). Mathematically, \(\lim\limits_{x \to a} f(x) = f(a)\). In this problem, continuity at \(x = 0\) requires that \(\lim\limits_{x \to 0} f(x) = f(0)\) and \(f(0) = c\).

Step 2: Simplify the Function for Limit Evaluation

Consider the limit \(\lim\limits_{x \to 0} \frac{9x - 3\sin(3x)}{5x^3}\). We notice that both the numerator and denominator approach 0 as \(x\) approaches 0, which suggests the use of L'Hôpital's rule to resolve the indeterminate form.

Step 3: Apply L'Hôpital's Rule

According to L'Hôpital's Rule, if a limit leads to an indeterminate form \(\frac{0}{0}\), we can differentiate the numerator and denominator separately and then evaluate the limit. Differentiate the numerator: \(\frac{d}{dx}(9x - 3\sin(3x)) = 9 - 9\cos(3x)\). Differentiate the denominator: \(\frac{d}{dx}(5x^3) = 15x^2\).

Step 4: Evaluate the Limit After First Differentiation

Evaluate the limit \(\lim\limits_{x \to 0} \frac{9 - 9\cos(3x)}{15x^2}\). This still results in an indeterminate form \(\frac{0}{0}\), so apply L'Hôpital's Rule a second time.

Step 5: Apply L'Hôpital's Rule Again

Differentiate the numerator and the denominator for the second time. Differentiating the numerator gives \(-27\sin(3x)\) and differentiating the denominator gives \(30x\).

Step 6: Evaluate the Second Limit

Evaluate the limit \(\lim\limits_{x \to 0} \frac{-27\sin(3x)}{30x}\). This results in \(\lim\limits_{x \to 0} \frac{-27 \cdot 3\cos(3x)}{30}\). Since \(\cos(0) = 1\), the limit simplifies to \(-\frac{27 \cdot 3}{30} = -\frac{81}{30} = -\frac{27}{10}\).

Step 7: Compare Limit and Function Value at Zero

For the function to be continuous at \(x = 0\), we need the limit found in the previous step to match the function value at \(x = 0\), which is \(c\). Thus, set \(c = -\frac{27}{10}\).

Key Concepts

Limit Evaluation

When dealing with functions involving limits, especially when you want to ensure continuity at a specific point, limit evaluation becomes crucial. Evaluating a limit means determining what value a function approaches as the independent variable approaches a certain value.
In this exercise, we need to evaluate the limit of the function: \[\lim\limits_{x \to 0} \frac{9x - 3\sin(3x)}{5x^3}\] The first thing we observe is that as \(x\) approaches 0, both the numerator \(9x - 3\sin(3x)\) and the denominator \(5x^3\) approach 0, resembling the fraction \(\frac{0}{0}\), which is an indeterminate form.
This evaluative process helps us understand how the function behaves near \(x = 0\). To accurately find this limit, we turn to a powerful calculus method called L'Hôpital's Rule, which helps resolve indeterminate forms.

Indeterminate Forms

Indeterminate forms, such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), arise when directly substituting a value into a function results in an ambiguous form. These forms don't provide clear answers, which is why we use different calculus techniques to resolve them.
In our function, we observed the indeterminate form \(\frac{0}{0}\) when substituting \(x = 0\). Specifically, for \[\lim\limits_{x \to 0} \frac{9x - 3\sin(3x)}{5x^3}\]the expression yields \(0/0\) because both the numerator and denominator independently tend toward zero as \(x\) approaches zero. Dealing with such forms requires a systematic approach to ascertain the true behavior of the function near that point.
Techniques such as simplification, factorization, or more advanced methods like L'Hôpital's Rule provide means to query the underlying limits accurately, resolving these intriguing forms into concrete numbers.

L'Hôpital's Rule

L'Hôpital's Rule is an essential tool in calculus for handling limits that produce indeterminate forms. The rule states that if the limit of a quotient results in \[\frac{0}{0} \text{ or } \frac{\infty}{\infty}\] you can find the limit of the derivatives of the numerator and denominator. In our exercise, we evaluated \[\lim\limits_{x \to 0} \frac{9x - 3\sin(3x)}{5x^3}\] Initially, a direct substitution gave \(\frac{0}{0}\). Thus, we took derivatives:

• Numerator: \(9 - 9\cos(3x)\)
• Denominator: \(15x^2\)

Re-evaluating the limit after a first differentiation also resulted in \(\frac{0}{0}\). So, we applied L'Hôpital's Rule again, taking a second set of derivatives:

• Numerator: \(-27\sin(3x)\)
• Denominator: \(30x\)

Finally evaluating the limit of this new expression resolved the indeterminate form, yielding \(-\frac{27}{10}\). This value was crucial to aligning the limit with the constant \(c\) to ensure continuity at \(x=0\). L'Hôpital's Rule thus not only allowed us to compute the limit but also facilitated the function's continuity condition by equating it to \(c\).