Problem 33 - 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

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

Step 1: Define Continuity at x=0

For the function to be continuous at \(x=0\), \(\lim_{{x \to 0}} f(x) = f(0)\). This implies \(\lim_{{x \to 0}} f(x) = c\). We need to find the limit \(\lim_{{x \to 0}} f(x)\) for \(x eq 0\).

Step 2: Substitute and Simplify the Limit

We have the function \(f(x) = \frac{9x - 3\sin 3x}{5x^3}\). To find the limit as \(x\) approaches 0, consider:\[\lim_{{x \to 0}} \frac{9x - 3\sin 3x}{5x^3}\]Use the expansion \(\sin 3x \approx 3x - \frac{(3x)^3}{6} + \cdots\) for small \(x\). The first term is nullified \(9x - 3(3x) = 0\), and the dominant term is \(-\frac{1}{2}(3x)^3\).

Step 3: Apply L'Hopital's Rule or Simplify Further

Since both the numerator and the denominator approach 0, apply L'Hôpital's Rule or continue simplifying by comparing the highest remaining order terms in Taylor expansion:\[\lim_{{x \to 0}} \frac{-\frac{(3x)^3}{2}}{5x^3} = \lim_{{x \to 0}} \frac{-27x^3/2}{5x^3} = \lim_{{x \to 0}} \frac{-27}{10} = -\frac{27}{10}\]Thus, the limit is \(-\frac{27}{10}\).

Step 4: Set Continuity Condition

For continuity at \(x=0\), set the calculated limit equal to the function value at \(x=0\), which implies:\[c = \lim_{{x \to 0}} f(x) = -\frac{27}{10}\]Thus, \(c = -\frac{27}{10}\) makes the function continuous at \(x=0\).

Step 5: Verify Continuity

Ensure that substituting \(c = -\frac{27}{10}\) into the function retains continuity. The left-hand limit, right-hand limit, and value at \(x=0\) are all equal, confirming our value of \(c\) achieves continuity.

Key Concepts

Continuity at a point

Continuity at a point is a fundamental concept in calculus and serves as the foundation for understanding continuous functions. For a function to be continuous at a specific point, the limit of the function as it approaches that point must be equal to the function's value at the point itself.

In mathematical terms, if we want function \( f(x) \) to be continuous at \( x = a \), we must ensure:

• \( \lim_{{x \to a}} f(x) \) exists.
• \( f(a) \) is defined.
• \( \lim_{{x \to a}} f(x) = f(a) \).

For the function in question, we were asked to find a value of \( c \) such that \( f(x) \) is continuous at \( x = 0 \). By calculating the limit \( \lim_{{x \to 0}} f(x) \) and setting it equal to \( c \), we make \( f(x) \) continuous at that point. Ensuring these conditions are met confirms that the function behaves predictably and smoothly at \( x = 0 \).

L'Hôpital's Rule

L'Hôpital's Rule is a tremendously useful tool in calculus for computing limits that initially appear to be indeterminate, such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). When faced with such situations, L'Hôpital's Rule allows us to take the derivative of the numerator and the derivative of the denominator and then re-evaluate the limit.

For example, if \( \lim_{{x \to c}} \frac{f(x)}{g(x)} \) presents an indeterminate form of \( \frac{0}{0} \), we apply:

• \( \lim_{{x \to c}} \frac{f'(x)}{g'(x)} \), provided this limit exists.

In our exercise, although we simplified directly using a series expansion, applying L'Hôpital's Rule could have been another approach if needed. It underscores the importance of having multiple strategies in tackling calculus problems, particularly when faced with complex limit evaluations.

Taylor expansion

Taylor expansion is a powerful mathematical tool that helps us approximate functions using polynomials around a specific point. It decomposes a function into an infinite sum of terms calculated from the values of its derivatives at a single point.

The Taylor series for a function \( f(x) \) centered at \( x = a \) is given by:

• \( f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \ldots \)

In our problem, we employed a Taylor expansion to approximate \( \sin 3x \) when \( x \) is close to zero. The small angle approximation tells us that \( \sin 3x \approx 3x - \frac{(3x)^3}{6} + \ldots \). By substituting this expansion into our function, we could simplify it, making it easier to evaluate the limit as \( x \) approaches zero. This clever technique helps break down seemingly complicated expressions into manageable parts.

Trigonometric limits

Understanding limits involving trigonometric functions is a crucial aspect of calculus, especially when working with small angles or when trigonometric expressions appear in complex functions. A foundational limit is \( \lim_{{x \to 0}} \frac{\sin x}{x} = 1 \). This result helps evaluate the behavior of sine functions as the angle approaches zero.

In our scenario, the expression \( 9x - 3\sin 3x \) in the numerator required approximating \( \sin 3x \) using its series expansion to better handle the limit. Recognizing and applying basic trigonometric limits as well as using expansions effectively allows us to simplify and solve problems that initially seem difficult at first glance.

These techniques often involve substituting approximations for complex terms, transforming them into algebraic forms that are easier to resolve. By doing so, we achieved a consistent approach to handling and solving limits involving trigonometry, reinforcing core calculus concepts.