Question

Removable and Nonremovable Discontinuities In Exercises \(35-60,\) find the \(x\) -values (if any) at which \(f\) is not continuous. Which of the discontinuities are removable? \ $$ f(x)=\left\\{\begin{array}{ll}{\csc \frac{\pi x}{6},} & {|x-3| \leq 2} \\\ {2,} & {|x-3|>2}\end{array}\right. $$

Short answer

This function has nonremovable discontinuities at \( x = 2n \), with n an integer such that \( 1 < n < 5 \), where the function \( f(x) = \csc(\frac{πx}{6}) \) has asymptotes. It has removable discontinuities at \( x = 1 \) and \( x = 5 \) if the function's left and right-hand limits aren't equal at these points.

Step-by-step solution

Determine the domain of the function

Firstly, let's take a look at the domains of the function's components. The function \( \csc(\frac{πx}{6}) \), which is the reciprocal of the sine function, is undefined for any value of \( x \) that makes \( \sin(\frac{πx}{6}) = 0 \). In this case, \( x = 2n \) where \( n \) is an integer, are points of discontinuity for the first function case because the denominator of \( \csc(\frac{πx}{6}) \) becomes zero. On the other hand, the function \( 2 \) is defined for any real number, so no points of discontinuity exist for the second case.

Identify the function's behavior at the bounds

We need to explore what happens to \( f(x) \) when \( x \) approaches \( 1 \) and \( 5 \) from both sides. These are the points where the function changes between the two cases. If the limits from the left and right are not equal, we have points of discontinuity. Thus, we need to check four limits: \( \lim_{{x \to 1^-}} f(x) \), \( \lim_{{x \to 1^+}} f(x) \), \( \lim_{{x \to 5^-}} f(x) \), and \( \lim_{{x \to 5^+}} f(x) \).

Evaluate the limits

Substitute the corresponding x-values to compute the limits. For \( x \to 1^- \) and \( x \to 5^+ \), we have \( f(x) = 2 \). For \( x \to 1^+ \) and \( x \to 5^- \), we substitute into \( \csc(\frac{πx}{6}) \). The main difficulty here is that \( \csc(\frac{πx}{6}) \) is undefined at \( x = 2n \) in the interval \( 1 < x < 5 \), but this doesn't happen at \( x = 1 \) and \( x = 5 \), so the limits can be computed normally.

Determine the removable and nonremovable discontinuities

The points \( x = 2n \), where \( n \) is integer in the interval \( 1 < x < 5 \), are nonremovable discontinuities because \( \csc(\frac{πx}{6}) \) has a vertical asymptote, and the value of \( f(x) \) can't be redefined to make the function continuous. On the other hand, if the limits computed in the previous step for \( x = 1 \) and \( x = 5 \) are different, they represent removable discontinuities, as we can redefine \( f(x) \) at these points.

Key concepts

Discontinuity in Functions

Understanding discontinuities in functions is crucial for students learning calculus. A discontinuity occurs when a function is not continuous at a point in its domain. This means that as we approach this point from either side, the function does not smoothly continue. Instead, there is a 'break' or 'jump' in the graph of the function.

There are several types of discontinuities, primarily removable and nonremovable. A removable discontinuity is present when a function has an undefined point or a hole at a certain value of x, but the limit exists as x approaches the value from both sides. This type of discontinuity can often be 'fixed' by redefining the value of the function at that point. On the other hand, a nonremovable discontinuity, also known as an essential discontinuity, occurs when the limits from the left and the right do not match, or at least one of them does not exist. Examples include vertical asymptotes or jumps in the function.

For the function given in the exercise, we analyze the type of discontinuity present by examining its behavior at specific points in its domain. This helps students understand where and why the function 'breaks' and provides insight into the behavior of mathematical models in real-world problems.

Domain of Function

The domain of a function encompasses all the input values (x-values) for which the function is defined. It is essentially the set of all possible 'x' that you can plug into the function without causing any undefined behavior (like dividing by zero, taking the square root of a negative number in real numbers, etcetera).

When dealing with the domain, students must consider the restrictions that different operations impose on a function. For example, the cosecant function, being the reciprocal of the sine function, will be undefined whenever the sine of an angle is zero — because you cannot divide by zero. This is why identifying the domain is the first step in finding discontinuities in the exercise provided. Once the domain is established, you can further analyze the behavior of the function within that domain to identify points of continuity or discontinuity.

Limits of a Function

The concept of limits in calculus is fundamental in understanding the behavior of functions, especially near points of discontinuity. A limit examines the behavior of a function as it approaches a certain value from either the left or the right side. If these two one-sided limits are equal, the two-sided limit exists, and that value is what the function is approaching.

When solving the provided exercise, computing the limits as x approaches the points where the function's definition changes is essential. This helps in determining whether the function gradually tends towards a specific value or if there's a sudden jump or gap — a hallmark of discontinuities. Limits play a critical role in defining continuity and are also used to calculate derivatives and integrals, which are the cornerstones of calculus.

Cosecant Function

The cosecant function, denoted as \(\csc(x)\), is the reciprocal of the sine function, meaning \(\csc(x) = \frac{1}{\sin(x)}\). It is defined for all real numbers except where the sine function is zero, since dividing by zero is undefined. For graphing purposes and analysis, this means the cosecant function has vertical asymptotes at these points.

Exploring the cosecant function in the context of the exercise helps students to see the application of these concepts in trigonometric functions. As the sine function oscillates between -1 and 1, its reciprocal will have no defined value at sine's zero crossings, giving rise to nonremovable discontinuities in the cosecant function. Recognizing when these occur is a key aspect of solving the problem at hand and further illustrates the behavior of trigonometric functions within their domains.