Question

In Exercises \(19-24,(\) a) find each point of discontinuity. (b) Which of the discontinuities are removable? not removable? Give reasons for your answers. $$f(x)=\left\\{\begin{array}{ll}{1-x^{2},} & {x \neq-1} \\ {2,} & {x=-1}\end{array}\right.$$

Short answer

The function \(f(x)\) is discontinuous at \(x=-1\). The discontinuity is removable because the limit of \(f(x)\) as \(x\) approaches -1 exists.

Step-by-step solution

Identify the points of discontinuity

The function \(f(x)\) is defined as \(1-x^{2}\) for all \(x\) except at \(x=-1\) where it is defined as \(2\). The function \(1-x^{2}\) is a continuous function for all \(x\), except potentially at \(x=-1\) where the definition changes. So, we only need to check for discontinuity at \(x=-1\).

Check for discontinuity at x = -1

This can be done by calculating the left and right-hand limits at \(x=-1\). If the two one-sided limits are equal, then the limit at that point exists. If the limit and the function's value at \(x=-1\) are equal, then the function is continuous at that point, otherwise, it is discontinuous. The left-hand limit as \(x\) approaches -1 is \(lim_{x \to -1^-}(1-x^{2}) = 0\). The right-hand limit as \(x\) approaches -1 is \(lim_{x \to -1^+}(1-x^{2}) = 0\). The value of \(f(x)\) at \(x=-1\) is \(2\). Since the value of \(f(x)\) at \(x=-1\) is not equal to the limit at \(x=-1\), the function is discontinuous at \(x=-1\).

Classify the discontinuity

Since the limit of \(f(x)\) as \(x\) approaches -1 exists, but is not equal to \(f(-1)\), the discontinuity at \(x=-1\) is removable.

Key concepts

Removable Discontinuity

Understanding the concept of a removable discontinuity is essential in the realm of calculus. It occurs when a function exhibits a gap at a particular value, but the limit exists. This means the function could be made continuous simply by redefining it at that point.

Let's illustrate this with an example from the mentioned exercise. We have a function defined as two different expressions depending on the value of x. At all points except where x is -1, the function behaves nicely and follows the rule given by the expression \(1 - x^2\). However, at \(x = -1\), the function suddenly jumps in value to 2, creating a gap.

This gap, or discontinuity, can be 'removed' by redefining the function at \(x = -1\) to match its limit at that point. Since the limit as x approaches -1 from either side is 0, and the function is defined to be 2 only at that specific point, this discontinuity is, in fact, removable by aligning the function’s value with the limit.

Limit of a Function

The limit of a function is a fundamental concept in calculus, encapsulating the idea of prediction. It answers the question: as x approaches a certain value, what does the function's output approach?

The textbook solution reveals that the left-hand limit (as x approaches from negative values) and the right-hand limit (as x approaches from positive values) both yield the same result, which is zero in this case. Notating this concept mathematically, we say that \( \text{lim}_{x \to -1^-} (1-x^2) = 0 \) and \( \text{lim}_{x \to -1^+} (1-x^2) = 0 \).

When these one-sided limits agree, it tells us that the function is approaching the same value from both sides of our point of interest. This makes the limit at that point clearly defined, even if the function itself might not match this value precisely at that point—as is the case in our textbook problem.

Continuity in Calculus

The notion of continuity is pivotal in calculus. A function is continuous at a point if you can draw the function at that value without lifting your pen. Formally, a function is continuous at a certain point if and only if three conditions are met: the function is defined at that point, the limit as x approaches that point exists, and the limit and the function's value at that point are the same.

Following the steps in our solution, we find that the function \(f(x)\) is indeed discontinuous at \(x = -1\). Although the function is defined at \(x = -1\) and the limit exists, the value of the function does not match the limit. Therefore, it breaks the rule for continuity. If the function had matched the limit, in this case, no discontinuity would have occurred. Such a nuance in the definition of continuity makes it an intriguing topic in the learning curve for students diving into calculus.