NOTE: You may view the original PDFs of the notes written in class by clicking on the link for the 1:30 Section or 2:30 Section, respectively.

GOAL: Be able to determine where and how a function misbehaves.

Lecture 4: Continuity

SIDENOTE: In the handwritten notes, "fcn" or "fn" is shorthand for "function."

Intuitively, a function is continuous if there are no abrupt changes in the graph.

Hand-drawn graph of a smooth wavy curve f crossing the x-axis, labeled Continuous
Figure 1
A hand-drawn sketch of coordinate axes with a smooth, unbroken wavy curve labeled f. The curve oscillates above and below the horizontal axis, crossing it several times without any breaks or gaps. A red arrow points to a red dot marked on the curve where it crosses the axis, indicating the curve passes smoothly through that point. Below the sketch is the handwritten label "Continuous".
Hand-drawn graph with a jump, a hole, and a vertical asymptote, labeled Discontinuous here
Figure 2
A hand-drawn sketch of coordinate axes showing a curve with several discontinuities. On the left, a line rises to an open circle, indicating a removable discontinuity, then a separate segment begins with a filled dot at a lower point and descends, showing a jump. Further right the curve dips, passes through an open circle, then rises steeply toward a dashed vertical asymptote; on the other side of the asymptote the curve comes down from above and flattens toward the horizontal axis. Three red arrows point upward to the locations of the jump, the hole, and the asymptote, joined by red lines to the red handwritten label "Discontinuous here" beneath the graph.

More precisely,

Definition: A function \( f \) is continuous at a point \( c \) if

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

Otherwise, \( f \) is said to be discontinuous at \( c \) and \( c \) is called a discontinuity.

Classifying Discontinuities

There are 3 ways a function can be discontinuous:

  1. \( \lim_{x \to c} f(x) \) DNE
  2. \( f(c) \) is undefined
  3. \( \lim_{x \to c} f(x) \ne f(c) \)

Example 1

Discuss the continuity of the Heaviside function \( H(x) = \begin{cases} 1 & x \ge 0 \\ 0 & x < 0 \end{cases} \) at \( x = 0 \)

Hand-drawn graph of the Heaviside step function with open circles at the origin
Figure 3
A hand-drawn sketch of coordinate axes in blue and red. A blue horizontal ray extends to the right at height 1, starting from a small open circle just above the origin on the vertical axis. A red horizontal ray extends to the left along the x-axis from an open circle at the origin. Together these depict the Heaviside step function, which equals 1 for x greater than or equal to 0 and 0 for x less than 0, with a jump at the origin.

Recall \( \lim\limits_{x \to 0} H(x) \) DNE since \( \lim\limits_{x \to 0^-} H(x) = 0 \), but \( \lim_{x \to 0^+} H(x) = 1 \). So by (1), \( H(x) \) is discontinuous at \( x = 0 \).

This is an example of a jump discontinuity; in a jump discontinuity, the left and right limits exist, but they are not equal. This causes a "jump" in the graph.

Example 2

Repeat for \( f(x) = \frac{x^2 - x - 2}{x - 2} \) at \( x = 2 \)

Plugging 2 into \(f\) gets us the indeterminate form \(\frac{0}{0}\), so \(f(2)\) is undefined. \( f \) is discontinuous at \( x = 2 \) by (2).

This is an example of a hole (or removable discontinuity) at \( x = 2\). This is when the limit exists, but something goes wrong with the function value. In this example, the limit:

\[ \lim_{x \to 2} \frac{x^2 - x - 2}{x - 2} = \lim_{x \to 2} \frac{(x+1)(x-2)}{x-2} = \lim_{x \to 2} (x+1) = 3 \]

But, \(f(2)\) is undefined causing the discontinuity

Hand-drawn graph of a line with an open circle at x equals 2, y equals 3
Figure 4
A hand-drawn red sketch of coordinate axes. A straight line rises from lower left to upper right, crossing the axes. The value 3 is marked on the vertical axis and the value 2 is marked on the horizontal axis. At the point where \( x = 2 \) and \( y = 3 \) there is an open circle on the line, indicating a hole in the graph at \( x = 2 \).

Question: Why is it called removable?

Answer: Define a new function:

\[ g(x) = \begin{cases} \frac{x^2 - x - 2}{x - 2} & x \ne 2 \\ 3 & x = 2 \end{cases} \]

where \(g(x) = f(x) = \frac{x^2 - x - 2}{x-2}\) everywhere except \(0\), while at \(0\), \(g(0) = 3 = \lim_{x \to 2} \frac{x^2 - x - 2}{x - 2}\). Assigning \(g(0)\) this way "plugs the hole" that \(f(x)\) has making \(g(x)\) continuous everywhere, as shown in Figure 5.

Hand-drawn graph of line g of x with a filled dot at x equals 2, y equals 3
Figure 5
A hand-drawn black sketch of coordinate axes with a straight line rising from lower left to upper right, labeled \( g(x) \). The value 3 is marked on the vertical axis and the value 2 is marked on the horizontal axis. At the point where \( x = 2 \) and \( y = 3 \) there is a solid filled dot on the line, showing that the hole has been filled in and the function is now continuous there.

Example 3

Repeat for \( f(x) = \begin{cases} 1 - x^2 & x \ne 0 \\ 2 & x = 0 \end{cases} \) at \( x = 0 \)

Hand-drawn downward parabola with open circle at 1 and filled dot at 2 on the y-axis
Figure 6
A hand-drawn red sketch of coordinate axes showing a downward-opening parabola \( y = 1 - x^2 \) whose branches cross the horizontal axis near \( x = -1 \) and \( x = 1 \) and continue downward with arrows. At the top of the parabola on the vertical axis there is an open circle at height 1, marked with a dashed line and the label 1. Above it, at height 2 on the vertical axis, there is a solid filled dot labeled 2, representing the defined value \( f(0) = 2 \).

\begin{align*} \lim_{x \to 0} f(x) =&~ \lim_{x \to 0} (1 - x^2) = 1 \\ f(0) =&~ 2 \end{align*}

Here we have \(\lim_{x \to 0} \neq f(0)\), so \(f\) is discontinuous at \(x=0\) by (3). Since the limit exists, \( f \) has a hole at \( x = 0 \).

Example 4 (Multiple Discontinuities)

Repeat for \( f(x) = \frac{x(x-5)}{x(x-1)} \)

Discontinuities could occur when the denominator is 0. I.e., when

\[ x(x-1) = 0 \]

Solving, we get that the possible locations of discontinuities are \(x=0\) and \(x=1\).

At \( x = 0 \):

\[ \lim_{x \to 0} \frac{x(x-5)}{x(x-1)} = \lim_{x \to 0} \frac{x-5}{x-1} = 5 \]

\( \Rightarrow f \) has a hole at \( x = 0 \)

At \( x = 1 \):

\( f(1) = \frac{-4}{0} \Rightarrow f(1) \) is undefined and neither the left nor right limit exists. This indicates it is NOT a jump discontinuity and instead shows \(f\) has a vertical asymptote at \(x=1\)

This is an example of an infinite discontinuity (another word for vertical asymptote). Here, at least one of the one-sided limits do not return a number (i.e., there are either \(\infty\) or \(-\infty\)).

Hand-drawn graph with vertical asymptote, open circle above the x-axis and hyperbola-like branches
Figure 7
A small hand-drawn sketch in red of the graph of the rational function. Horizontal and vertical axes are drawn. A vertical dashed line to the right of the y-axis marks the vertical asymptote at \( x = 1 \). An open circle is drawn on the y-axis above the origin, indicating the hole at \( x = 0 \) where the limit value is 5. To the left of the asymptote a curve rises steeply upward toward the asymptote from the upper region, and to the right of the asymptote a branch comes up from below and levels off approaching a horizontal level as \( x \) increases.

Properties of Continuous Functions

  1. We can define "left/right continuity" like what we did with limits. To be precise, we say a function \(f\) is left-continuous at a point \(c\) if: \[ \lim_{x \to c^-} f(x) = f(c) \] Similarly, a function is right-continuous at \(c\) if: \[ \lim_{x \to c^+} f(x) = f(c) \] A function is then continuous if it is both left and right continuous.
  2. It is common to say "\( f \) is continuous on an interval" to mean it is continuous at every point on that interval. To be precise, we say \(f(x)\) is continuous on an interval \(I\) if \(f(x)\) is continous at \(c\) for every point \(c\) in \(I\). For example, \(\ln x\) is continuous on \((0,\infty)\).
Hand-drawn graph of the natural logarithm curve crossing the x-axis at one
Figure 8
A small hand-drawn sketch in red showing horizontal and vertical axes with the curve of \( y = \ln x \). The curve comes up steeply from far below near the y-axis, crosses the x-axis just to the right of the origin, and then rises slowly and flattens out as \( x \) increases, illustrating that \( \ln x \) is continuous on \( (0, \infty) \).

Example 5

Discuss the continuity of \( f(x) = \begin{cases} e^{-x} & x < 0 \\ \sqrt{x} & x \ge 0 \end{cases} \)

Hand-drawn graph with decaying exponential branch left of origin and square root branch right of origin
Figure 9
A hand-drawn red sketch with horizontal and vertical axes. To the left of the y-axis a decreasing curve descends from the upper left toward the y-axis, ending at an open circle at height 1 on the y-axis, representing \( e^{-x} \) for \( x < 0 \). At the origin there is a filled dot, representing \( f(0) = 0 \). To the right of the origin a square-root shaped curve rises from the origin, increasing and flattening as \( x \) increases, representing \( \sqrt{x} \) for \( x \ge 0 \). The gap between the open circle at height 1 and the filled dot at 0 shows the jump at \( x = 0 \).

\[f(0) = 0 \]

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

\( \Rightarrow f \) is right-continuous at \(x=0\)

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \left( e^{-x} \right) = 1 \ne 0 = f(0) \]

\( \Rightarrow f \) is not left continuous at \(x=0\)

Therefore, \( f \) is continuous everywhere except \(0\). I.e., it is continuous on \( (-\infty, 0) \cup (0, \infty) \). As a reminder, \(\cup\) is the union symbol; think of it as the word "or." So, if \(x\) is in \((-\infty, 0)\) or \((0,\infty)\), then \(f(x)\) is continuous there.

Since the left and right limits exist, just unequal, there is a jump at \( x = 0 \).

Theorem

On their domain, polynomials, rational functions, root functions, trig functions, inverse trig functions, exponential functions, and logarithmic functions are continuous.

The "on their domain" part is important. For example, \(\frac{1}{x}\) is discontinuous at \(0\), but it is not in the domain. So, \(\frac{1}{x}\) is continuous everywhere else.

Hand-drawn graph of one over x with two hyperbola branches about the axes
Figure 10
A small freehand sketch in red showing horizontal and vertical axes crossing at the origin. Two hyperbola branches are drawn: one in the upper right quadrant approaching the horizontal axis from above and rising steeply near the vertical axis, and one in the lower left quadrant, mirrored through the origin. The curve is labeled with the fraction one over x, indicating the graph of \( \frac{1}{x} \), which has a vertical asymptote at the vertical axis and a horizontal asymptote along the horizontal axis.

Example 6 (Similar to HW4, #11)

\[ f(x) = \begin{cases} -6x - \frac{\pi}{2} & x \le -\frac{\pi}{2} \\ \cos x & -\frac{\pi}{2} < x \le \frac{\pi}{2} \\ 5\sin x + 9 & x > \frac{\pi}{2} \end{cases} \]

All functions are continuous on their parts; the candidates for discontinuities are where the formula changes. I.e., at \( x = -\frac{\pi}{2}, \frac{\pi}{2} \).

At \( x = -\frac{\pi}{2} \):

\[ \begin{gathered} \lim_{x \to -\frac{\pi}{2}^-} f(x) = \lim_{x \to -\frac{\pi}{2}^-} \left[ -6x - \frac{\pi}{2} \right] = 3\pi - \frac{\pi}{2} \\[0.5em] \lim_{x \to -\frac{\pi}{2}^+} f(x) = \lim_{x \to -\frac{\pi}{2}^+} \left[ \cos x \right] = \cos\left( \frac{-\pi}{2} \right) = 0 \end{gathered} \]

\( \Rightarrow f \) has a jump discontinuity at \( x = -\frac{\pi}{2} \)

Hand-drawn unit circle with axes and a marked point at the bottom
Figure 11
A quick red freehand sketch of a unit circle centered on a pair of crossing axes. A small filled dot is marked at the bottom of the circle, on the negative vertical axis, corresponding to the angle \( -\frac{\pi}{2} \). A small arrow near the right of the circle suggests the direction of rotation. This is used to evaluate \( \cos\left(-\frac{\pi}{2}\right) = 0 \).
Two hand-drawn reference right triangles for special angles
Figure 12
Two red freehand right triangles drawn side by side as trigonometric reference triangles. The left triangle has hypotenuse labeled 2, the angle at the left vertex labeled pi over 6, the angle at the top labeled pi over 3, the vertical side labeled 1, and the horizontal base labeled square root of 3. The right triangle has hypotenuse labeled square root of 2, the angle at the left vertex labeled pi over 4, the angle at the top labeled pi over 4, the vertical side labeled 1, the horizontal base labeled 1, and a small square marking the right angle at the bottom right corner.

At \( x = \frac{\pi}{2} \):

\[ \begin{gathered} \lim_{x \to \frac{\pi}{2}^-} f(x) = \lim_{x \to \frac{\pi}{2}^-} \left[ \cos x \right] = 0 \\[0.5em] \lim_{x \to \frac{\pi}{2}^+} f(x) = \lim_{x \to \frac{\pi}{2}^+} \left[ 5\sin x + 9 \right] = 5\sin\frac{\pi}{2} + 9 = 5 + 9 = 14 \end{gathered} \]

\( \Rightarrow f \) has a jump discontinuity at \( x = \frac{\pi}{2} \)

Compositions

Theorem Let \( f \) be continuous at \( x = G \), where \( G = \lim_{x \to c} g(x) \). Then,

\[ \lim_{x \to c} f(g(x)) = f\left( \lim_{x \to c} g(x) \right) = f(G) \]

In particular, if \( g \) is continuous at \( c \) and \( f \) is continuous at \( g(c) \), then the composition \( f(g(x)) \) is continous at \( c \).

Example 7(a)

\[ \lim_{x \to 1} \frac{x^2 (x-1)}{(x^2+3)(x-1)} = \lim_{x \to 1} \frac{x^2}{x^2+3} = \frac{1}{1+3} = \frac{1}{4} \]

Example 7(b)

\[ \lim_{x \to 1} \sqrt{\frac{x^2(x-1)}{(x^2+3)(x-1)}} \;\underset{\substack{\uparrow \\ \sqrt{\;}\ \text{is} \\ \text{cont. at } \frac{1}{4}}}{=}\; \sqrt{\lim_{x \to 1} \frac{x^2(x-1)}{(x^2+3)(x-1)}} = \sqrt{\frac{1}{4}} = \frac{1}{2} \]

The moral here is we can shove limits inside continuous functions, which can be helpful to solve some limits.

Example 8

\[ \lim_{x \to c} e^{f(x)} = e^{\lim_{x \to c} f(x)} \]

(Non-Examinable) Roots of Continuous Functions

This will not show up on homeworks/exams; you need not worry about this. This is just highlighting an application that's not in the standard curriculum.

One of the important properties of continuous functions for applications is we are able to find decimal approximations for their roots. In particular,

Theorem: There is a finite process that has a continuous function \(f\), and an interval containing at least one root, as input and returns a decimal approximation of the root up to a specified error \(\varepsilon\).

The discussion on how to do this is a bit long, but a typical Calculus I student should be able to understand it with a bit of gusto. The details can be found here.