Problem 79 - Thomas Calculus

Question

Continuous extension 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

c = \frac{27}{10} ensures continuity at \( x = 0 \).

Step 1: Identify the Limit Condition for Continuity

For the function \( f(x) \) to be continuous at \( x = 0 \), the limit of \( f(x) \) as \( x \) approaches 0 must equal \( f(0) \). In mathematical terms, this means \( \lim_{{x \to 0}} \frac{9x - 3\sin 3x}{5x^3} = c \).

Step 2: Simplify Using L'Hôpital's Rule

The expression \( \frac{9x - 3\sin 3x}{5x^3} \) gives an indeterminate form \( \frac{0}{0} \) as \( x \to 0 \). We apply L'Hôpital's Rule, which allows us to take derivatives of the numerator and denominator. So, differentiate both to get \( \frac{d}{dx}[9x - 3\sin 3x] = 9 - 9\cos 3x \) and \( \frac{d}{dx}[5x^3] = 15x^2 \).

Step 3: Apply L'Hôpital's Rule Until Detachment

Apply L'Hôpital's Rule again to \( \frac{9 - 9\cos 3x}{15x^2} \). This produces another indeterminate form. Differentiating again, the numerator becomes \( 27\sin 3x \) and the denominator becomes \( 30x \).

Step 4: Evaluate the Limit After Sufficient Derivatives

The new expression is \( \frac{27\sin 3x}{30x} \). Applying L'Hôpital's Rule one last time, differentiate to get \( 81\cos 3x \) and \( 30 \), leading to \( \lim_{{x \to 0}} \frac{81\cos 3x}{30} = \frac{81 \cdot 1}{30} = \frac{27}{10} \).

Step 5: Determine the Required Value of c

Since \( \lim_{{x \to 0}} \frac{9x - 3\sin 3x}{5x^3} = \frac{27}{10} \) to satisfy continuity, set \( c = \frac{27}{10} \). Thus, the value of \( c \) that makes \( f(x) \) continuous at \( x = 0 \) is \( c = \frac{27}{10} \).

Step 6: Conclusion: Justify the Choice of c

The choice \( c = \frac{27}{10} \) ensures the limit condition for continuity is met: \( \lim_{{x \to 0}} f(x) = f(0) \). As calculated, both the limit and \( f(0) \) are \( \frac{27}{10} \), thus the function is continuous at \( x = 0 \).

Key Concepts

L'Hôpital's Rule

L'Hôpital's Rule is an essential tool for dealing with limits that are in an indeterminate form, such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). This rule enables us to differentiate the numerator and the denominator of a function separately to simplify the expression. In the context of our problem, we needed to find

• the limit of \( \frac{9x - 3\sin 3x}{5x^3} \) as \( x \) approaches zero.
• All terms in the expression become zero, indicating an indeterminate form \( \frac{0}{0} \).

So, we apply L'Hôpital's Rule, differentiate the numerator and denominator, and simplify step-by-step. Repeated application of L'Hôpital's Rule transforms the expression:

• First differentiation: transform into \( \frac{9 - 9\cos 3x}{15x^2} \).
• Second: becomes \( \frac{27\sin 3x}{30x} \).
• Finally: into a simple \( \frac{81\cos 3x}{30} \).

When evaluated at \( x = 0 \), this yields \( \frac{27}{10} \). By continuously differentiating, we leave the indeterminate form, thus simplifying calculation by making the limits more straightforward. In the end, the rule helps us answer the original continuity question by calculating \( c \).

Limits

The concept of limits is central to calculus and continuity. It helps us understand what happens to a function as the input gets closer to a point. In our exercise, determining if

• the limit of the given function \( f(x) \) as \( x \rightarrow 0 \)
• exists, and equates to \( f(0) \).

It was crucial in asserting that the function is continuous at that point. When evaluating a limit, especially when a function leads to an indeterminate form like \( \frac{0}{0} \), techniques like L'Hôpital's Rule become invaluable.The main steps involve:

• substituting into the function, checking for continuity or indeterminacy.
• if indeterminate, applying calculus techniques to reduce complexity.

Ultimately, the limit tells us the behavior of \( f(x) \) near zero and helps us find the condition where function equals its value at the critical point, \( x = 0 \). The resolved limit for this function was \( \frac{27}{10} \), establishing it as the correct value for \( c \) to maintain continuity.

Continuous Functions

A function is continuous at a point if the limit as the variable approaches the point equals the function’s value at that point. For the piecewise function given in our problem,

• we need the value of \( c \) to ensure continuity at \( x = 0 \).
• In mathematical terms, this meant solving: \( \lim_{{x \to 0}} f(x) = f(0) = c \).

For continuity, both \( \lim_{{x \to 0}} f(x) \) and \( f(0) \) should consistently equal each other. In our exercise, we showed continuity through the involvement of

• calculations using L'Hôpital's Rule, ensuring the limit seamlessly met the function's definition at the specified point of interest.
• Resulting in the need to set \( c = \frac{27}{10} \).

Why is this important? Because continuous functions have no gaps or jumps at the point, you are evaluating. This means that smoothly moving from one value to the next without sudden changes is significant in calculus, providing a foundation for integral and derivative operations.