2.7: Continuity

Section Preview

Note to the Instructor (click to expand)

Before we get to derivatives, we have one last limit-related concept—continuity. For students at this level, the most important result in this section is that, as long as we have a function that is continuous at \(x = a\), we can simply substitute \(a\) when evaluating the limit of the function as \(x\) is approaching \(a\). Don't skip over the Intermediate Value Theorem—it appears throughout the rest of the course!

Core Prerequisites (i.e., "Understanding the Lesson"): The following prerequisite topics (which have not already been listed as prerequisites in previous sections of this text) are required to understand the core concepts in this lesson.

(click to expand)

• An Overview of Functions
  • Function Composition and Domain: The Composite Function Theorem and the proof of trigonometric continuity require understanding nested functions \(f\big(g(x)\big)\); Example 8 treats \(\cos\left(x-\tfrac{\pi}{2}\right)\) as a composition.

Operational Prerequisites (i.e., "Completing the Homework"): The following prerequisite topics (which have not already been listed as prerequisites in previous sections of this text) are needed to complete the homework.

(click to expand)

• Trigonometry
  • Evaluating Trigonometric Functions at Special Angles: The IVT application (Example 9) and Example 8 require exact values such as \(\cos0=1\), \(\cos\tfrac{\pi}{2}=0\), and \(\sin\pi=0\) (Checkpoint 8).
  • Trigonometric Identities — Sum and Difference of Angles Identities: The proof that trigonometric functions are continuous expands \(\cos\big((x-a)+a\big)\) as \(\cos(x-a)\cos a-\sin(x-a)\sin a\).

The following is a list of learning objectives for this section.

Learning Objectives (click to expand)

• Explain the three conditions for continuity at a point.
• Describe three kinds of discontinuities.
• Define continuity on an interval.
• Evaluate limits using the Generalized Direct Substitution Property.
• State the theorem for limits of composite functions.
• Understand and investigate uses of the Intermediate Value Theorem.

We begin our investigation of continuity by exploring what it means for a function to have continuity at a point. Intuitively, a function is continuous at a particular point if there is no break in its graph at that point; however, as often is the case, intuition has its flaws.

Continuity at a Point

Before we formally define what it means to be continuous at a point, let's consider various functions that fail to meet our intuitive notion of what it means to be continuous at a point. We then create a list of conditions that prevent such failures.

Our first function of interest is shown in Figure \(\PageIndex{1}\). We see that the graph of \(f(x)\) has a hole at \(a\). In fact, \(f(a)\) is undefined. At the very least, for \(f(x)\) to be continuous at \(a\), we need the following condition:

i. \(f(a)\) is defined

Figure \(\PageIndex{1}\): The function \(f(x)\) is not continuous at \(a\) because \(f(a)\) is undefined.

However, as we see in Figure \(\PageIndex{2}\), this condition alone is insufficient to guarantee continuity at the point \(a\). Although \(f(a)\) is defined, the function has a gap at \(a\). In this example, the gap exists because \(\displaystyle \lim_{x \to a}f(x)\) does not exist. We must add another condition for continuity at \(a\)—namely,

ii. \(\displaystyle \lim_{x \to a}f(x)\) exists

Figure \(\PageIndex{2}\): The function \(f(x)\) is not continuous at \(a\) because \(\displaystyle \lim_{x \to a}f(x)\) does not exist.

However, as we see in Figure \(\PageIndex{3}\), these two conditions by themselves do not guarantee continuity at a point. The function in this figure satisfies both of our first two conditions but is still not continuous at \(a\). We must add a third condition to our list:

iii. \(\displaystyle \lim_{x \to a}f(x)=f(a)\)

Figure \(\PageIndex{3}\): The function \(f(x)\) is not continuous at \(a\) because \(\displaystyle \lim_{x \to a}f(x) \neq f(a)\).

Now, we put our list of conditions together and form a definition of continuity at a point.

Definition: Continuous at a Point

A function \(f(x)\) is said to be continuous at a point \(a\) if each of the following three conditions is satisfied:

1. \(f(a)\) is defined
2. \(\displaystyle \lim_{x \to a}f(x)\) exists
3. \(\displaystyle \lim_{x \to a}f(x)=f(a)\)

A function is discontinuous at a point \(a\)if it fails to be continuous at \(a\).

Interactive Element: When Is a Function Continuous at a Point?

The following Interactive Element might help you visualize the third condition for continuity at a point.

Interact: Drag the red point up and down the line \(x=2\).

Observation: \(f(2)\) is defined and the limit is \(3\) throughout, yet the graph is continuous only when the point lands in the hole — where \(f(2)=3=\lim_{x\to 2}f(x)\). The value and the limit must be equal.

The next three examples demonstrate how to apply this definition to determine whether a function is continuous at a given point. These examples illustrate situations in which each of the conditions for continuity in the definition succeeds or fails.

Example \(\PageIndex{1A}\): Determining Continuity at a Point, Condition 1

Using the definition, determine whether the function \(f(x)=\frac{x^2-4}{x-2}\) is continuous at \(x=2\). Justify the conclusion.

Solution

Let's begin by trying to calculate \(f(2)\). We can see that \(f(2) = \frac{4 - 4}{2 - 2}\), which is undefined. Therefore, \(f(x)=\frac{x^2-4}{x-2}\) is discontinuous at \(2\) because \(f(2)\) is undefined. The graph of \(f(x)\) is shown in Figure \(\PageIndex{4}\).

Figure \(\PageIndex{4}\): The function \(f(x)\) is discontinuous at \(2\) because \(f(2)\) is undefined.

Example \(\PageIndex{1B}\): Determining Continuity at a Point, Condition 2

Using the definition, determine whether the function\[f(x)=\begin{cases}-x^2+4, & \mathrm{if} \; x \leq 3 \\ 4x-8, & \mathrm{if} \; x>3\end{cases}\nonumber\]is continuous at \(x=3\). Justify the conclusion.

Solution

Let's begin by trying to calculate \(f(3)\).\[f(3)=-(3^2)+4=-5.\nonumber\]Thus, \(f(3)\) is defined. Next, we calculate \(\displaystyle \lim_{x \to 3}f(x)\). To do this, we must compute \(\displaystyle \lim_{x \to 3^-}f(x)\) and \(\displaystyle \lim_{x \to 3^+}f(x)\):\[\displaystyle \lim_{x \to 3^-}f(x)=-(3^2)+4=-5\nonumber\]and\[\displaystyle \lim_{x \to 3^+}f(x)=4(3)-8=4.\nonumber\]Therefore, \(\displaystyle \lim_{x \to 3}f(x)\) does not exist. Thus, \(f(x)\) is not continuous at \(3\). The graph of \(f(x)\) is shown in Figure \(\PageIndex{5}\).

Figure \(\PageIndex{5}\): The function \(f(x)\) is not continuous at \(3\) because \(\displaystyle \lim_{x \to 3}f(x)\) does not exist.

Example \(\PageIndex{1C}\): Determining Continuity at a Point, Condition 3

Using the definition, determine whether the function\[f(x)=\begin{cases}\frac{\sin x}{x}, & \text{if } x \neq 0\\1, & \text{if } x=0\end{cases}\nonumber\]is continuous at \(x=0\).

Solution

First, observe that\[f(0)=1.\nonumber\]Next,\[\displaystyle \lim_{x \to 0}f(x)=\lim_{x \to 0}\frac{\sin x}{x}=1.\nonumber\]Last, compare \(f(0)\) and \(\displaystyle \lim_{x \to 0}f(x)\). We see that\[\displaystyle f(0)=1=\lim_{x \to 0}f(x).\nonumber\]Since all three of the conditions in the definition of continuity are satisfied, \(f(x)\) is continuous at \(x=0\).

Checkpoint \(\PageIndex{1}\)

Using the definition, determine whether the function\[f(x)=\begin{cases}2x+1, & \text{if }x<1\\2, & \text{if }x=1\\ -x+4, & \text{if }x>1\end{cases}\nonumber\]is continuous at \(x=1\). If the function is not continuous at \(1\), indicate the condition for continuity at a point that fails to hold.

Answer

\(f\) is not continuous at \(1\) because \(\displaystyle f(1)=2 \neq 3=\lim_{x \to 1}f(x)\).

By applying the definition of continuity and previously established theorems concerning the evaluation of limits, we can state the following theorem.

Theorem: Continuity of Polynomials and Rational Functions

Polynomials and rational functions are continuous at every point in their domains.

Proof

Previously, we showed that if \(p(x)\) and \(q(x)\) are polynomials, \(\displaystyle \lim_{x \to a}p(x)=p(a)\) for every polynomial \(p(x)\) and \(\displaystyle \lim_{x \to a}\frac{p(x)}{q(x)}=\frac{p(a)}{q(a)}\) as long as \(q(a) \neq 0\). Therefore, polynomials and rational functions are continuous on their domains.

Q.E.D.

We now apply this theorem to determine the points at which a given rational function is continuous.

Example \(\PageIndex{2}\): Continuity of a Rational Function

For what values of \(x\) is \(f(x)=\frac{x+1}{x-5}\) continuous?

Solution

The rational function \(f(x)=\frac{x+1}{x-5}\) is continuous for every value of \(x\) except \(x=5\).

Checkpoint \(\PageIndex{2}\)

For what values of \(x\) is \(f(x)=3x^4-4x^2\) continuous?

Answer

\(f(x)\) is continuous at every real number.

Types of Discontinuities

As we have seen in Example \(\PageIndex{1A}\) and Example \(\PageIndex{1B}\), discontinuities take on several different appearances. We classify the types of discontinuities we have seen thus far as removable, infinite, or jump discontinuities. Intuitively, a removable discontinuity is a discontinuity in which the graph has a hole. A jump discontinuity is a non-infinite discontinuity for which the sections of the function do not meet up. Finally, an infinite discontinuity is a discontinuity located at a vertical asymptote. Figure \(\PageIndex{6}\) illustrates the differences in these types of discontinuities. Although these terms provide a handy way of describing three common types of discontinuities, remember that not all discontinuities fit neatly into these categories.

Figure \(\PageIndex{6}\): Discontinuities are classified as (a) removable, (b) jump, or (c) infinite.

These three discontinuities are formally defined as follows:

Definition: Types of Discontinuities

If \(f(x)\) is discontinuous at \(a,\) then

1. \(f\) has a removable discontinuity at \(a\) if \(\displaystyle \lim_{x \to a}f(x)\) exists. (Note: When we state that \(\displaystyle \lim_{x \to a}f(x)\) exists, we mean that \(\displaystyle \lim_{x \to a}f(x)=L\), where \(L\) is a real number.)
2. \(f\) has a jump discontinuity at \(a\) if \(\displaystyle \lim_{x \to a^-}f(x)\) and \(\displaystyle \lim_{x \to a^+}f(x)\) both exist, but \(\displaystyle \lim_{x \to a^-}f(x) \neq \lim_{x \to a^+}f(x)\). (Note: When we state that \(\displaystyle \lim_{x \to a^-}f(x)\) and \(\displaystyle \lim_{x \to a^+}f(x)\) both exist, we mean that both are real-valued and that neither takes on the values \(\pm \infty\).)
3. \(f\) has an infinite discontinuity at \(a\) if \(\displaystyle \lim_{x \to a^-}f(x)= \pm \infty\) or \(\displaystyle \lim_{x \to a^+}f(x)= \pm \infty\).

Example \(\PageIndex{3}\): Classifying a Discontinuity

In Example \(\PageIndex{1A}\), we showed that \(f(x)=\frac{x^2-4}{x-2}\) is discontinuous at \(x=2\). Classify this discontinuity as removable, jump, or infinite.

Solution

To classify the discontinuity at \(2\) we must evaluate \(\displaystyle \lim_{x \to 2}f(x)\):\[\begin{array}{rcl} \displaystyle \lim_{x \to 2}f(x) & = & \displaystyle \lim_{x \to 2}\frac{x^2-4}{x-2}\\[8pt] & = & \displaystyle \lim_{x \to 2}\frac{(x-2)(x+2)}{x-2}\\[8pt] &= & \displaystyle \lim_{x \to 2}(x+2)\\[8pt] & = & 4 \\ \end{array}\nonumber\]Since \(f\) is discontinuous at \(2\) and \(\displaystyle \lim_{x \to 2}f(x)\) exists, \(f\) has a removable discontinuity at \(x=2\).

Example \(\PageIndex{4}\): Classifying a Discontinuity

In Example \(\PageIndex{1B}\), we showed that\[f(x)=\begin{cases}-x^2+4, &\text{if }x \leq 3\\4x-8, &\text{if }x>3\end{cases}\nonumber\]is discontinuous at \(x=3\). Classify this discontinuity as removable, jump, or infinite.

Solution

Earlier, we showed that \(f\) is discontinuous at \(3\) because \(\displaystyle \lim_{x \to 3}f(x)\) does not exist. However, since \(\displaystyle \lim_{x \to 3^-}f(x)=-5\) and \(\displaystyle \lim_{x \to 3^+}f(x)=4\) both exist, we conclude that the function has a jump discontinuity at \(3\).

Example \(\PageIndex{5}\): Classifying a Discontinuity

Determine whether \(f(x)=\frac{x+2}{x+1}\) is continuous at \(-1\). If the function is discontinuous at \(-1\), classify the discontinuity as removable, jump, or infinite.

Solution

The function value \(f(-1)\) is undefined. Therefore, the function is not continuous at \(-1\). To determine the type of discontinuity, we must determine the limit at \(-1\). We see that \(\displaystyle \lim_{x \to -1^-}\frac{x+2}{x+1}=-\infty\) and \(\displaystyle \lim_{x \to -1^+}\frac{x+2}{x+1}=+\infty\). Therefore, the function has an infinite discontinuity at \(-1\).

Checkpoint \(\PageIndex{5}\)

For\[f(x)=\begin{cases}x^2, &\text{if }x \neq 1\\3, & \text{if }x=1\end{cases}\nonumber\]decide whether \(f\) is continuous at \(1\). If \(f\) is not continuous at \(1\), classify the discontinuity as removable, jump, or infinite.

Answer

Discontinuous at \(1\); removable

Continuity over an Interval

Now that we have explored the concept of continuity at a point, we extend that idea to continuity over an interval. As we develop this idea for different types of intervals, it may be useful to keep in mind the intuitive (but flawed) idea that a function is continuous over an interval if we can use a pencil to trace the function between any two points in the interval without lifting the pencil from the paper. In preparation for defining continuity on an interval, we begin by defining what it means for a function to be continuous from the right at a point and continuous from the left at a point.

Definition: Continuity from the Right and from the Left

A function \(f(x)\) is said to be continuous from the right at \(a\) if \(\displaystyle \lim_{x \to a^+}f(x)=f(a)\).

A function \(f(x)\) is said to be continuous from the left at \(a\) if \(\displaystyle \lim_{x \to a^-}f(x)=f(a)\).

Definition: Continuity over Intervals

A function is continuous over an open interval if it is continuous at every point in the interval. A function \(f(x)\) is continuous over a closed interval of the form \([a,b]\) if it is continuous at every point in \((a,b)\) and is continuous from the right at \(a\) and is continuous from the left at \(b.\) Analogously, a function \(f(x)\) is continuous over an interval of the form \((a,b]\) if it is continuous over \((a,b)\) and is continuous from the left at \(b.\) Continuity over other types of intervals is defined similarly.

Requiring that \(\displaystyle \lim_{x \to a^+}f(x)=f(a)\) and \(\displaystyle \lim_{x \to b^-}f(x)=f(b)\) ensures that we can trace the graph of the function from the point \((a,f(a))\) to the point \((b,f(b))\) without lifting the pencil. If, for example, \(\displaystyle \lim_{x \to a^+}f(x) \neq f(a)\), we would need to lift our pencil to jump from \(f(a)\) to the graph of the rest of the function over \((a,b]\).

Example \(\PageIndex{6}\): Continuity on an Interval

State the interval(s) over which the function \(f(x)=\frac{x-1}{x^2+2x}\) is continuous.

Solution

Since \(f(x)=\frac{x-1}{x^2+2x}\) is a rational function, it is continuous at every point in its domain. The domain of \(f(x)\) is the set \((-\infty,-2) \cup (-2,0) \cup (0,+\infty)\). Thus, \(f(x)\) is continuous over each of the intervals \((-\infty,-2),(-2,0)\), and \((0,+\infty)\).

Example \(\PageIndex{7}\): Continuity over an Interval

State the interval(s) over which the function \(f(x)=\sqrt{4-x^2}\) is continuous.

Solution

From the Limit Laws, we know that \(\displaystyle \lim_{x \to a}\sqrt{4-x^2}=\sqrt{4-a^2}\) for all values of \(a\) in \((-2,2)\). We also know that \(\displaystyle \lim_{x \to -2^+}\sqrt{4-x^2}=0\) exists and \(\displaystyle \lim_{x \to 2^-}\sqrt{4-x^2}=0\) exists. Therefore, \(f(x)\) is continuous over the interval \([-2,2]\).

Checkpoint \(\PageIndex{7}\)

State the interval(s) over which the function \(f(x)=\sqrt{x+3}\) is continuous.

Answer

\([-3,+\infty)\)

Continuity and Compositions

Theorem: Composite Function Theorem

If \(f(x)\) is continuous at \(L\) and \(\displaystyle \lim_{x \to a}g(x)=L\), then\[\displaystyle \lim_{x \to a}f\big(g(x)\big)=f\big(\lim_{x \to a}g(x)\big)=f(L).\nonumber\]

The Composite Function Theorem allows us to expand our ability to compute limits. In particular, this theorem ultimately allows us to demonstrate that trigonometric functions are continuous over their domains.

Before we move on to Example \(\PageIndex{8}\), recall that earlier, in the section on Limit Laws, we showed \(\displaystyle \lim_{x \to 0}\cos x=1=\cos(0)\). Consequently, we know that \(f(x)=\cos x\) is continuous at \(0\). In Example \(\PageIndex{8}\), we see how to combine this result with the Composite Function Theorem.

Example \(\PageIndex{8}\): Limit of a Composite Cosine Function

Evaluate \(\displaystyle \lim_{x \to \pi /2}\cos\left(x-\frac{\pi}{2}\right)\).

Solution

The given function is a composite of \(\cos x\) and \(x-\frac{\pi}{2}\). Since \(\displaystyle \lim_{x \to \pi /2}\left(x-\frac{\pi}{2}\right)=0\) and \(\cos x\) is continuous at \(0\), we may apply the Composite Function Theorem. Thus,\[\displaystyle \lim_{x \to \pi /2}\cos\left(x-\frac{\pi}{2}\right)=\cos\left(\lim_{x \to \pi /2}\left(x-\frac{\pi}{2}\right)\right)=\cos(0)=1.\nonumber\]

Checkpoint \(\PageIndex{8}\)

Evaluate \(\displaystyle \lim_{x \to \pi}\sin(x- \pi )\).

Answer

\(0\)

The proof of the next theorem uses the Composite Function Theorem as well as the continuity of \(f(x)=\sin x\) and \(g(x)=\cos x\) at the point \(0\) to show that trigonometric functions are continuous over their entire domains.

Theorem: Continuity of Trigonometric Functions

Trigonometric functions are continuous over their entire domains.

Proof

We begin by demonstrating that \(\cos x\) is continuous at every real number. To do this, we must show that \(\displaystyle \lim_{x \to a}\cos x=\cos a\) for all values of \(a\).\[\displaystyle \begin{array}{rclcl} \displaystyle \lim_{x \to a}\cos x & = & \displaystyle \lim_{x \to a}\cos((x-a)+a) & \quad & \left(\text{rewrite }x=x-a+a\right) \\ & = & \displaystyle \lim_{x \to a}(\cos(x-a)\cos a-\sin(x-a)\sin a) & \quad & \left(\text{Sum and Difference of Angles Identities} \right) \\ & = & \displaystyle \cos(\lim_{x \to a}(x-a))\cos a-\sin(\lim_{x \to a}(x-a))\sin a & \quad & \left(\text{since }\lim_{x \to a}(x-a)=0,\text{ and }\sin x\text{ and }\cos x\text{ are continuous at }0 \right) \\ & = & \cos(0)\cos a-\sin(0)\sin a & & \\ & = & 1 \cdot \cos a-0 \cdot \sin a & & \\ & = & \cos a & & \\ \end{array}\nonumber\]The proof that \(\sin x\) is continuous at every real number is analogous. Because the remaining trigonometric functions may be expressed in terms of \(\sin x\) and \(\cos x\), their continuity follows from the quotient limit law.

Q.E.D.

As you can see, the Composite Function Theorem is invaluable in demonstrating the continuity of trigonometric functions. As we continue our study of calculus, we revisit this theorem many times.

The Intermediate Value Theorem

Functions that are continuous over intervals of the form \([a,b]\), where \(a\) and \(b\) are real numbers, exhibit many useful properties. Throughout our study of calculus, we will encounter many powerful theorems concerning such functions. The first of these theorems is the Intermediate Value Theorem.

Theorem: Intermediate Value Theorem (IVT)

Let \(f\) be continuous over a closed, bounded interval \([a,b]\). If \(z\) is any real number between \(f(a)\) and \(f(b)\), then there is a number \(c\) in \([a,b]\) satisfying \(f(c)=z\).

See Figure \(\PageIndex{7}\) for a visualization of the IVT.

Figure \(\PageIndex{7}\): There is a number \(c \in [a,b]\) that satisfies \(f(c)=z\).

Interactive Element: The Intermediate Value Theorem

The following Interactive Element might help you visualize why a continuous function on a closed interval must hit every height between its endpoint values.

Interact: Drag the red point up and down the left edge to choose a target height \(z\) between \(f(1)=1\) and \(f(5)=5\).

Observation: No matter which \(z\) you pick in that band, the level line always meets the curve — there is always at least one \(c\) in \([1,5]\) with \(f(c)=z\).

Example \(\PageIndex{9}\): Application of the Intermediate Value Theorem

Show that \(f(x)=x-\cos x\) has at least one zero.

Solution

Since \(f(x)=x-\cos x\) is continuous over \((-\infty,+\infty)\), it is continuous over any closed interval of the form \([a,b]\). If you can find an interval \([a,b]\) such that \(f(a)\) and \(f(b)\) have opposite signs, you can use the Intermediate Value Theorem to conclude there must be a real number \(c\) in \((a,b)\) that satisfies \(f(c)=0\). Note that\[f(0)=0-\cos(0)=-1<0\nonumber\]and\[f\left(\dfrac{\pi}{2}\right) = \dfrac{\pi}{2}-\cos\dfrac{\pi}{2} = \dfrac{\pi}{2} > 0.\nonumber\]Using the Intermediate Value Theorem, we can see that there must be a real number \(c\) in \([0, \pi /2]\) that satisfies \(f(c)=0\). Therefore, \(f(x)=x-\cos x\) has at least one zero.

Example \(\PageIndex{10}\): When Can You Apply the Intermediate Value Theorem?

If \(f(x)\) is continuous over \([0,2],f(0)>0\) and \(f(2)>0\), can we use the Intermediate Value Theorem to conclude that \(f(x)\) has no zeros in the interval \([0,2]\)? Explain.

Solution

No. The Intermediate Value Theorem only allows us to conclude that we can find a value between \(f(0)\) and \(f(2)\); it doesn't allow us to conclude that we can't find other values. To see this more clearly, consider the function \(f(x)=(x-1)^2\). It satisfies \(f(0)=1>0,f(2)=1>0\), and \(f(1)=0\).

Example \(\PageIndex{11}\): When Can You Apply the Intermediate Value Theorem?

For \(f(x)=1/x,f(-1)=-1<0\) and \(f(1)=1>0\). Can we conclude that \(f(x)\) has a zero in the interval \([-1,1]\)?

Solution

No. The function is not continuous over \([-1,1]\). The Intermediate Value Theorem does not apply here.

Checkpoint \(\PageIndex{11}\)

Show that \(f(x)=x^3-x^2-3x+1\) has a zero over the interval \([0,1]\).

Answer

\(f(0)=1>0,\;f(1)=-2<0;\;f(x)\) is continuous over \([0,1]\). It must have a zero on this interval.

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Success in College: Cheating and Consequences

The following is an excerpt from an article at WorldNetDaily.com by Dennis Prager.

A report by Donald McCabe, professor of organization management at Rutgers University and founding president of the Center for Academic Integrity at Duke University, describes a pretty bad ethical situation in America's high schools. Seventy-five percent of students engage in "serious cheating" (such as repeated cheating and significant plagiarism).

It gets worse.

First, according to the professor, this epidemic is the culmination of three decades of increased cheating in high schools.

Second, about half of all students (not just half of all cheaters) see nothing wrong with cheating.

As one who has written and lectured on ethics for 30 years, during which time I, too, have spoken at high schools across America, I am certain that these statistics are accurate. I also believe that I know the major reasons.

One reason is that parents emphasize many things over character.

A second reason is the absurd emphasis on getting into the best college. This preoccupation on the part of parents (which often begins in preschool) has utterly distorted our children's (and high schools') priorities. Because students are taught that getting into the right college is the most important thing in their lives, they readily compromise their character to achieve this goal. And they could not care less about learning. Since the purpose of school is to get good grades, learning is incidental, and loving learning is a non-issue.

A third and final reason is the death of the concept of sin, a product of the thoroughly secular education American high schools offer. As noted, far worse than the number of students who cheat is the number of students who think that there is nothing wrong with cheating. I can testify from my talks at high schools that the vast majority of students believe what they have been taught - that right and wrong - are entirely subjective concepts. There is no morality higher than one's own; therefore, right and wrong are defined by the individual, who can rationalize any behavior.

When students do not understand a concept, they need help. Instead, some students in this situation choose to cheat. In college, any form of cheating is "academic dishonesty."

1. Has your professor discussed or posted their cheating policy?
2. What happens if you cheat in this math class?

To be clear, cheating - in any form - results in more than just a "zero" on that assignment. This includes, but is not limited to, using notes on exams, using a device that is not allowed during an exam, or using unapproved technology to help during an exam. If I suspect a student of cheating, I will have them meet me outside of class time in my office to discuss what happened and why it is considered cheating. They will receive a zero on the assignment/exam and will be reported to the Student Discipline Officer for further discussions. The Student Discipline Officer is involved for two reasons: to give the student tools to help them break the "cheating habit," and to create an official record within the Los Rios Community College District to track that student's cheating offenses - if multiple offenses have occurred, the student could like be placed on academic probation or, in severe cases, completely expelled from the District.

- Professor Simpson, Cosumnes River College

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Success in Mathematics: Holding Onto Old, Broken Techniques

It's common for students to learn a problem-solving technique years ago and not want to learn a different method because they are comfortable with their old habits. Unfortunately, we often fail to recognize that if our old, comfortable technique worked well, we should be much further along in our mathematical journey.

This adherence to comfort is widespread for all human beings in various situations. Whether it's a dead-end job, an abusive relationship, or a bad habit, we tend to ignore signals—remaining in a known situation rather than taking the chance to improve and evolve.

In general, when your professor gives you advice or a different approach to doing something, they are doing so from the lens of years of experience seeing students repeat old, broken techniques or habits that tend to lead students to make mistakes. Paying attention to these words of wisdom is critical.

EXAMPLE

Simplify the compound rational expression\[ \dfrac{-\frac{2}{(x+h)^2} + \frac{2}{x^2}}{h} \nonumber\]Students stuck in old habits tend to find the common denominator of the fractions in the numerator, make equivalent fractions, add fractions, divide the result by \(h\), then simplify. This entire process looks like the following:\[ \begin{array}{rcl} \dfrac{-\frac{2}{(x+h)^2} + \frac{2}{x^2}}{h} & = & \dfrac{-\frac{2x^2}{x^2(x+h)^2} + \frac{2(x+h)^2}{x^2(x+h)^2}}{h} \\[6pt] & = & \dfrac{\frac{-2x^2 + 2(x+h)^2}{x^2(x+h)^2}}{h} \\[6pt] & = & \dfrac{-2x^2 + 2(x+h)^2}{x^2(x+h)^2} \div h \\[6pt] & = & \dfrac{-2x^2 + 2(x+h)^2}{x^2(x+h)^2} \cdot \dfrac{1}{h} \\[6pt] & = & \dfrac{-2x^2 + 2(x+h)^2}{h x^2(x+h)^2} \\[6pt] & = & \dfrac{-2x^2 + 2(x^2+2hx + h^2)}{h x^2(x+h)^2} \\[6pt] & = & \dfrac{-2x^2 + 2x^2+4hx + 2h^2}{h x^2(x+h)^2} \\[6pt] & = & \dfrac{4hx + 2h^2}{h x^2(x+h)^2} \\[6pt] & = & \dfrac{h(4x + 2h)}{h x^2(x+h)^2} \\[6pt] & = & \dfrac{\cancel{h}(4x + 2h)}{\cancel{h} x^2(x+h)^2} \\[6pt] & = & \dfrac{(4x + 2h)}{ x^2(x+h)^2} \\[6pt] \end{array} \nonumber \]Compare this to the advice given by your math professor to begin by multiplying both numerator and denominator of the original expression by the LCD of all minor fractions within the compound rational expression.\[ \begin{array}{rcl} \dfrac{-\frac{2}{(x+h)^2} + \frac{2}{x^2}}{h} & = & \dfrac{-2x^2 + 2(x+h)^2}{hx^2(x+h)^2} \\[6pt] & = & \dfrac{-2x^2 + 2(x^2+2hx + h^2)}{h x^2(x+h)^2} \\[6pt] & = & \dfrac{-2x^2 + 2x^2+4hx + 2h^2}{h x^2(x+h)^2} \\[6pt] & = & \dfrac{4hx + 2h^2}{h x^2(x+h)^2} \\[6pt] & = & \dfrac{h(4x + 2h)}{h x^2(x+h)^2} \\[6pt] & = & \dfrac{\cancel{h}(4x + 2h)}{\cancel{h} x^2(x+h)^2} \\[6pt] & = & \dfrac{(4x + 2h)}{ x^2(x+h)^2} \\ \end{array} \nonumber \]As you can see, the second method is cleaner and less complex.

The moral of the story is, when learning, try to be open to... learn.