Problem 42 - Contemporary Precalculus

Question

A function \(f\) that is not defined at \(x=c\) is said to have \(a\) removable discontinuity at \(x=c\) if there is a function \(g\) such that \(g(c)\) is defined, \(g\) is continuous at \(x=c,\) and \(g(x)=f(x)\) for \(x \neq c .\) In Exercises \(39-43,\) show that the function \(f\) has a removable discontinuity by finding an appropriate function g. $$f(x)=\frac{\sin x}{x} \quad[\text {Hint}: \text { See Example 4 on pages } 882-883 .]$$

Step-by-Step Solution

Answer

Answer: The function \(g(x)\) is defined as: $$g(x) = \begin{cases} \frac{\sin x}{x}, & \text{if} \ x\neq0 \\ 1, & \text{if} \ x=0 \end{cases}$$

Step 1: Identify the point of discontinuity

The function \(f(x) = \frac{\sin x}{x}\) is not defined at \(x = 0\), so we'll focus on verifying if the function has a removable discontinuity at \(x = 0\).

Step 2: Finding the limit at the discontinuous point

We will find the limit of the function \(f(x)\) as \(x\) approaches \(0\). By applying L'Hopital's rule, we'll check if the limit exists and if finding it will help us find an appropriate function \(g(x)\) for the discontinuity of \(f(x)\). Consider the limit: $$\lim_{x \to 0} \frac{\sin x}{x}$$ Applying L'Hopital's rule, we get: $$\lim_{x \to 0} \frac{d(\sin x)}{dx} \cdot \lim_{x \to 0} \frac{1}{dx}$$ Now, the derivatives of numerator and denominator are: $$\frac{d(\sin x)}{dx} = \cos x \quad \text{and} \quad \frac{d(x)}{dx} =1$$ Thus, we get the limit as: $$\lim_{x \to 0} \frac{\cos x}{1} = \cos 0 = 1$$

Step 3: Define the function g(x)

Now that we have found the limit of \(f(x)\) as \(x\) approaches \(0\), we can use this value to define our function \(g(x)\). Given that the limit exists and is equal to \(1\), we can define \(g(x)\) as follows: $$g(x) = \begin{cases} \frac{\sin x}{x}, & \text{if} \ x\neq0 \\ 1, & \text{if} \ x=0 \end{cases}$$

Step 4: Verify g(x) is continuous at x = 0

We should now verify that the function \(g(x)\) is continuous at the point \(x=0\): $$\lim_{x \to 0} g(x) = \lim_{x \to 0} \frac{\sin x}{x} = 1$$ Since \(\lim_{x \to 0} g(x) = g(0)\), we can conclude that the function \(g(x)\) is continuous at \(x=0\).

Step 5: Conclusion

We have shown that the function \(f(x) = \frac{\sin x}{x}\) has a removable discontinuity at \(x = 0\), by finding an appropriate function \(g(x)\), which is continuous at \(x = 0\) and equal to \(f(x)\) for \(x \neq 0\).

Key Concepts

continuous function

A continuous function is one where small changes in the input (x-values) result in small changes in the output (f-values). For a function to be continuous at a point, there are three key conditions that must be satisfied:

• The function must be defined at the point.
• The limit of the function as it approaches that point from both directions must exist.
• The limit at the point must equal the function's value at that point.

In mathematical terms, a function \(f(x)\) is continuous at \(x = c\) if \( \lim_{x \to c} f(x) = f(c) \) and \(f(c)\) is defined.
This means if you draw the function's graph, you should be able to draw it without lifting your pencil. In our exercise, while \(f(x) = \frac{\sin x}{x}\) is not defined at \(x = 0\), we found an alternative function \(g(x)\), which made it continuous at that point by assigning the limit value at \(x = 0\).
This resolves the removable discontinuity by redefining the function only at the discontinuous point.

L'Hopital's rule

L'Hopital's rule is a useful method in calculus for finding the limit of indeterminate forms. Indeterminate forms are expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) that do not have a straightforward limit.
The rule states that if \( \lim_{x \to c} f(x) = \lim_{x \to c} g(x) = 0\) or \(\pm \infty\), then
\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \]
provided that this latter limit exists.

• Use L'Hopital's Rule when you encounter expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
• Differentiate the numerator and the denominator separately, then calculate the limit again.

In our exercise, L'Hopital's rule was applied to solve \( \lim_{x \to 0} \frac{\sin x}{x} \). Since \( \frac{0}{0} \) is an indeterminate form at \(x = 0\), we used its derivatives: \(\frac{d}{dx} \sin x = \cos x\) and \(\frac{d}{dx} x = 1\). The resulting limit \( \lim_{x \to 0} \frac{\cos x}{1} = 1\) confirmed \(g(x)\) is suitable for continuity.

limit

The concept of a limit is fundamental in calculus. Limits help describe the behavior of functions as they approach a specific point. More formally, the limit of a function \(f(x)\) as \(x\) approaches \(c\) is expressed as \( \lim_{x \to c} f(x) \). This determines what value \(f(x)\) gets arbitrarily close to as \(x\) gets nearer and nearer to \(c\).
This is particularly useful in studying functions at points where they are not directly defined, like points of discontinuity.

• If the limit of \(f(x)\) as \(x\) approaches \(c\) exists, and equals \(f(c)\), \(f(x)\) is continuous at \(x = c\).
• If not, the function could either have a removable discontinuity to be managed by redefining the function, or a more complex form of discontinuity.

In our scenario, although the function \(f(x) = \frac{\sin x}{x}\) is not defined at \(x = 0\), using the limit \( \lim_{x \to 0} \frac{\sin x}{x} = 1\), which we found using L'Hopital's Rule, helped redefine the function \(g(x)\) to make it continuous.