Continuity: Videos & Practice Problems

Continuity in mathematics refers to the behavior of a function at a specific point. A function is considered continuous at a point c if the limit of the function as x approaches c equals the function's value at that point. In mathematical terms, this can be expressed as:

$$\lim_{{x \to c}} f(x) = f(c)$$

When both sides of this equation are equal, the function is continuous at c. Conversely, if the limit does not equal the function value, the function is discontinuous at that point. Discontinuities can arise from various situations, such as holes, jumps, or asymptotes in the graph of the function.

To visually assess continuity, one can trace the graph of the function. If you can trace through a point without lifting your pencil, the function is continuous at that point. If you have to lift your pencil, it indicates a discontinuity. For example, if a function has a hole at x = 2, the limit may approach a certain value, but the function value at that point is undefined, indicating discontinuity.

Discontinuities can be classified into different types. A hole occurs when a function is undefined at a point where the limit exists. A jump discontinuity happens when the left-hand limit and right-hand limit at a point are not equal. Asymptotic discontinuities occur when the function approaches infinity or negative infinity at a certain point.

To determine continuity at specific values, one can evaluate both the limit and the function value. For instance, consider the following examples:

1. For c = -2: If the limit as x approaches -2 is 1 and the function value at -2 is also 1, the function is continuous at this point.

2. For c = 4: If the left-hand limit approaches 1 and the right-hand limit approaches 3, the limit does not exist, indicating a jump discontinuity since the function value at 4 is 1.

3. For c = 1: If the limit approaches negative infinity and the function value is undefined due to an asymptote, the function is discontinuous at this point.

Understanding these concepts of continuity and discontinuity is essential for analyzing functions and their behaviors in calculus. By practicing these evaluations, one can become proficient in identifying and classifying the continuity of various functions.

Understanding the continuity of a function is crucial in calculus, especially when you are not provided with a graph. Discontinuities can arise in various forms, such as jumps, holes, or asymptotes, particularly in rational and piecewise functions. To identify where a function is discontinuous, we can follow a systematic approach.

For rational functions, the key to finding discontinuities lies in the denominator. A rational function is expressed as the ratio of two polynomials, and it is continuous everywhere except where the denominator equals zero. For example, consider the function:

$$f(x) = \frac{x^2 - x - 6}{x + 2}$$

Here, the denominator is \(x + 2\). Setting this equal to zero gives:

$$x + 2 = 0 \implies x = -2$$

This indicates a potential discontinuity at \(x = -2\). To determine the nature of this discontinuity, we can check if the numerator can be factored to cancel with the denominator. If it can, we have a removable discontinuity (a hole); if not, we have an asymptote. In this case, the numerator factors to \((x - 3)(x + 2)\), allowing us to cancel the \((x + 2)\) term, confirming a removable discontinuity at \(x = -2\).

Next, we examine piecewise functions, which can also exhibit discontinuities at the points where the pieces meet. For instance, consider the piecewise function:

$$f(x) = \begin{cases} 4 & \text{if } x < -1 \\ x + 1 & \text{if } x \geq -1 \end{cases}$$

To check for discontinuity, we focus on the point where the pieces connect, which is \(x = -1\). We need to evaluate the limit as \(x\) approaches \(-1\) from both sides:

1. **Left-sided limit**: As \(x\) approaches \(-1\) from the left, \(f(x) = 4\), so:

$$\lim_{x \to -1^-} f(x) = 4$$

2. **Right-sided limit**: As \(x\) approaches \(-1\) from the right, \(f(x) = x + 1\), thus:

$$\lim_{x \to -1^+} f(x) = -1 + 1 = 0$$

Since the left-sided limit (4) and the right-sided limit (0) are not equal, the overall limit does not exist at \(x = -1\). Additionally, the function value at this point is:

$$f(-1) = -1 + 1 = 0$$

Comparing the limits and the function value, we see that they do not match, confirming a discontinuity at \(x = -1\). This type of discontinuity is classified as a jump discontinuity, as the function value jumps from 4 to 0.

In summary, to determine where a function is discontinuous, analyze rational functions by setting the denominator to zero and checking for removable discontinuities through factoring. For piecewise functions, evaluate the limits at the points where the pieces meet to identify any jumps or holes. This systematic approach will enhance your understanding of function continuity and discontinuity.

To determine the points of discontinuity for the function \( f(x) = \frac{x - 3}{x^2 + 2x - 15} \), we need to analyze the denominator, as discontinuities in rational functions occur where the denominator equals zero. Thus, we set the denominator \( x^2 + 2x - 15 \) to zero.

To solve this quadratic equation, we can factor it. We are looking for two numbers that multiply to \(-15\) (the constant term) and add to \(2\) (the coefficient of the linear term). The numbers that satisfy these conditions are \(-3\) and \(5\). Therefore, we can factor the quadratic as:

\( (x - 3)(x + 5) = 0 \)

Setting each factor equal to zero gives us the potential points of discontinuity:

\( x - 3 = 0 \) leads to \( x = 3 \)

\( x + 5 = 0 \) leads to \( x = -5 \)

Thus, the function is discontinuous at \( x = 3 \) and \( x = -5 \).

Next, we can classify the types of discontinuities. The factor \( x - 3 \) in the numerator cancels with the factor \( x - 3 \) in the denominator, indicating that \( x = 3 \) is a removable discontinuity (or a hole) in the graph of the function. Conversely, the factor \( x + 5 \) in the denominator does not cancel with any factor in the numerator, indicating that \( x = -5 \) is a vertical asymptote.

In summary, the function \( f(x) \) has two points of discontinuity: a removable discontinuity at \( x = 3 \) and a vertical asymptote at \( x = -5 \).

To determine the points of discontinuity in a piecewise function, we need to analyze the transition points between the different pieces of the function. In this case, we focus on two critical values: \( x = -2 \) and \( x = 0 \).

Starting with \( x = -2 \), we evaluate the left-sided and right-sided limits. The left-sided limit as \( x \) approaches \(-2\) is derived from the first piece of the function, yielding:

\[\lim_{{x \to -2^-}} f(x) = 5\]

For the right-sided limit, we use the second piece of the function, which is defined as \( x^2 + 2x + 1 \). Plugging in \(-2\) gives:

\[\lim_{{x \to -2^+}} f(x) = (-2)^2 + 2(-2) + 1 = 4 - 4 + 1 = 1\]

Since the left-sided limit (5) does not equal the right-sided limit (1), the overall limit as \( x \) approaches \(-2\) does not exist. Consequently, we find that:

\[\lim_{{x \to -2}} f(x) \neq f(-2)\]

Thus, there is a discontinuity at \( x = -2 \).

Next, we examine \( x = 0 \). Again, we calculate the left-sided and right-sided limits. The left-sided limit as \( x \) approaches \( 0 \) is:

\[\lim_{{x \to 0^-}} f(x) = 0^2 + 2(0) + 1 = 1\]

For the right-sided limit, we use the last piece of the function, which is \( x + 1 \). Evaluating at \( 0 \) gives:

\[\lim_{{x \to 0^+}} f(x) = 0 + 1 = 1\]

Since both the left-sided and right-sided limits are equal, we conclude:

\[\lim_{{x \to 0}} f(x) = 1\]

Now, we check the function value at \( x = 0 \):

\[f(0) = 0^2 + 2(0) + 1 = 1\]

Since the limit equals the function value, the function is continuous at \( x = 0 \).

In summary, the only point of discontinuity in this piecewise function occurs at \( x = -2 \), where the limits do not match. The function is continuous at \( x = 0 \).