# Continuity

Lesson Features »

Lesson Priority: Normal

Calculus $\longrightarrow$
Functions and Limits $\longrightarrow$

Objectives
• Understand continuity at a point or over an interval conceptually and visually based on graphs
• Define continuity of a function at a point using limits
• Define continuity of a function over an interval using limits
• Classify types of discontinuities, either based on graphs or based on function definitions
Lesson Description

We've talked in the past in Algebra Two and Pre-Calculus about a loose definition of what it means for a function to be continuous. Now we'll talk about a rigorous definition, using limits.

Practice Problems

Practice problems and worksheet coming soon!

## It's Good to Be Well-Connected

Continuity is a concept that could be studied and understood in Algebra, but is more acutely defined in here in Calculus, through the use of limits. So what exactly is Continuity all about? In short, it's a measure of whether a function is connected with itself. We'll see that this is easier to define visually / graphically than it is to define algebraically, though we will need to be experts on both for typical exam questions.Let's digest this concept first with a loose definition and a focus on the graphical approach, before moving to the more rigorous and tedious algebra based tasks.

## A Practical Definition

Functions are continuous along the parts of their graph that you can draw without picking up your pencil, and discontinuous at the place(s) where you had to pick up your pencil. This idea is the loose definition of continuity.Here is a graph of a function that has a point of discontinuity, followed by a graph that is continuous everywhere on the function's domain:  Note that $f(x)$ has a gap when $x=3$, and how it would not be possible to draw this graph without picking up your pencil. This is not a rigorous definition of continuity, but it is true that $f(x)$ is not continuous at $x=3$. It is also true that $f(x)$ is continuous everywhere else that we can tell. This leads is to our first important point about continuity:Continuity is not a yes or no question - i.e. a function is not either continuous or discontinuous identically. A function is continuous on the intervals that are connected, and discontinuous at the points or intervals on the graph that we would have had to pick up our pencil when drawing.

## What's the Real Question?

Let's take a look at a few more graphs to better understand what continuity and discontinuity each look like. This graph has two discontinuities - at $x=-1$ there is a hole in the graph, and at $x=2$ there is a jump in the graph. The question we're often going to need to answer is - "where is the graph continuous"? Since most parts of most graphs you will see are continuous, this question is best answered by taking the perspective that the function is continuous everywhere except for the places we identify that are discontinuous. So the graph above looks to be continuous everywhere except for $x=-1$ and $x=2$. If we are asked to present our answer in interval notation, then we would say $x$ is continuous on the interval $(-\infty, -1) \,\, \bigcup \,\, (-1, 2) \,\, \bigcup \,\, [2, \infty)$.Here's another one: While much of this graph is smooth and connected, there are a few places that stand out as "break points" or as places where, when drawing the function, we had to lift our pencil up off the page. The function is discontinuous between $x=-1$ and $x=2$, not just at the specific locations of $x=-1$ and $x=2$, because the function is not defined at all on this interval. Therefore, when asked, we would say that this function is continuous on the interval $(-\infty, -1) \,\, \bigcup \,\, (2,5) \,\, \bigcup \,\, (5, \infty)$.

## The Four Flavors of Discontinuity

Hopefully with these few examples, it is becoming clear how to visually determine the places or intervals on a function that are continuous, and those that are discontinuous. Before we look at the limit definition of how to determine continuity, let's specify the four categories of discontinuities that we will encounter.Jump DiscontinuitiesThe first type of discontinuity is called a jump discontinuity. This is the classic "gap" situation where, at the point of discontinuity, each side of the function is approaching different values. Note that in terms of continuity at that point of separation, it doesn't matter which side has the solid dot where the function is actually defined, nor does it matter if neither is solid - a discontinuity will exist at this point regardless. This type of discontinuity is typically exclusive to piecewise defined functions.Removable DiscontinuitiesThe next type of discontinuity is called a removable discontinuity. Both of the following similar functions exhibit this type of discontinuity:  The name comes from the idea that the function would be otherwise continuous in the vicinity of the point, were we to "plug the hole". In other words, each side of the function approaches the same value, but the function is not continuous because for whatever reason, the function value at that point is defined to be a different value, or not defined at that point at all (the first and second graphs above, respectively). The function is discontinuous either way, since the function has a puncture in it. Typically this happens naturally in rational functions that have a common factor in both the numerator and denominator (see the lesson on limits with zero in the denominator »).Infinite DiscontinuitiesUp third, we see one of the more common situations that occur in functions that we work with often. An infinite discontinuity occurs any time a vertical asymptote appears. This discontinuity category applies regardless of whether the graph approaches the same or different directions on each side of the asymptote. Both asymptotes in the following figure are infinite discontinuities. Typically, this type of behavior can appear naturally in rational functions that have a division by zero domain error. However, along with many of the stranger graphs in math, it can be a product of a piecewise defined function. In fact, it's possible to define a function piecewise such that one side of an asymptote has infinite behavior, while the other side has finite behavior. Here's an example of that usual case: Infinite Oscillation DiscontinuitiesLast but not least, we have one of the least frequent occurrences in functions we work with, as this issue is specific to trig functions. An infinitely oscillating discontinuity occurs when a periodic function oscillates with increasing frequency that becomes infinitely frequent at a specific instant. An example would be $f(x)=\sin(1/x)$ at $x=0$: Of all the discontinuity types, this one is easily the most unique, but it is also infrequent. Some teachers don't even cover it in fact, just to save time and energy.Note: while it is important and more palatable to digest the differences among these four classifications visually, we will also need to look at how to tell these four categories apart without graphs, using only the algebraic limit definition that we will turn to next. First we'll acquaint ourselves with the limit defintion approach, and then we will quickly revisit each of these four situations and see how to tell them apart with the limit method.
Pro Tip Many teachers explicitly test your ability to classify discontinuities into one of these four categories. On a test is is possible to be asked to do this using the graph approach that we are using now, but it is even more likely that if you are asked to do this task, that you will need to use the limit definition approach that we are about to look at.

## The Proper Approach

While most students find it easier to describe continuity with a visual approach, we must ultimately define it symbolically, using limits.
Define: Function ContinuityLet $f(x)$ be a function with domain $D$. We say that $f(x)$ is continuous on every point in its domain such that$$\lim_{x \to a} f(x) = f(a)$$(1)for all $a \in D$, and only if $\lim_{x \to a} f(x)$ exists and is finite.
A word interpretation of this definition is to say that a function is continuous at all places where the function does what it says it will do, remembering the all-important idea that limits have to do solely with function behavior. If the limit at a point does not exist for any reason or is not finite, then the point fails this limit definition and therefore a discontinuity exists at this point.Let's take another pass through each of the four types and see how each is defined using the limit definition of continuity.Jump Discontinuities - Limit DefinitionLet's take another look at a function with a single jump discontinuity, and apply the limit definition. The limit as $x$ approaches $2$ from the left looks to be $4$, but the limit as $x$ approaches $2$ from the right looks to be $1$. Since the one sided limits do not agree, the limit of the function at $x=2$ does not exist, and therefore we know immediately that the function is discontinuous at $x=2$.Look at same function at any other point on the graph, and notice how limit does always equal the function value. This reinforces the idea that the easiest way to describe continuity is to say that the function is continuous on its domain except for the places that you can identify discontinuities, which in this case is $x=2$.
Define: Jump DiscontinuityIf, at any point $a$ on $f(x)$$\lim_{x \to a^-} f(x) = c_1$$$$\lim_{x \to a^+} f(x) = c_2$$where$c_1$and$c_2$are finite, and$c_1 \neq c_2$, then there exists a jump discontinuity in the function at$x = a$.Stated differently, when you examine a point of potential discontinuity, and you find that each of the one-sided limits are finite but do not agree, then you know specifically that you have a jump discontinuity. Removable Discontinuities - Limit DefinitionRecall that a removable discontinuity was the case when the function looked nearly continuous but has a puncture or hole, whether or not the function is defined at the point of interest: or In terms of limits, the hallmark quality of a removable discontinuity is that each one side limit will agree, but the function value will not be equal to the limit. In the example above, we can see that$\lim_{x \to 3^{+}}$is$2$,$\lim_{x \to 3^{-}}$is$2$, but$f(3) \ne 2$. Define: Removable DiscontinuityIf, at any point on$f(x)$$\lim_{x \to a^-} f(x) = c_1$$$$\lim_{x \to a^+} f(x) = c_1$$$$f(a) \neq c_1$$where $c_1$ is finite, then there exists a removable discontinuity at in the function at $x=a$.In other words, any point where the two-sided limit exists but the function value evaluates to a different number is a place where the function will have a removable discontinuity.
Infinite Discontinuities - Limit DefinitionAs we've seen, the presence of any vertical asymptote creates an infinite discontinuity. If you find that either of the one-side limits is $+\infty$ or $-\infty$, then you know you're dealing with an infinite discontinuity (which is why they are named that way).
Define: Infinite DiscontinuityIf, at any point on $f(x)$$\lim_{x \to a^-} f(x) = \infty \,\, \mathrm{or} \,\, -\infty$$or$$\lim_{x \to a^+} f(x) = \infty \,\, \mathrm{or} \,\, -\infty$$then there exists an infinite discontinuity at in the function at$x=a$.In other words, any limit with an infinite result is an indicator that an infinite discontinuity exists. Oscillating Discontinuities - Limit DefinitionFinally, the oscillating discontinuity's hallmark is that the$\sin$or$\cos$function argument will approach infinity. For example, we looked at the function$y = \sin(1/x)$, which looks like The problem comes from the fact that as$x$approaches zero, the argument of the sine function approaches infinity. Since sine functions oscillate back and forth forever, no matter how far right you go, this behavior means that the function$\sin(1/x)$oscillates back and forth infinitely at$0$. The limit doesn't exist. Define: Oscillating DiscontinuityIf$f(x)$is a sinusoid, then an infinite oscillation will occur at any finite point$a$that makes the argument of the trig function approach infinity as$x$approaches$a$. Again, these infinite oscillations are distinguished from other types of discontinuities, and also uncommon. Remember! It is very possible that you are going to be asked to classify discontinuities without having a picture to guide you. Make sure you understand the limit definition, and can tell each of the four categories apart by their unique limit properties. Pro Tip If you can read the following table fluently and understand what it is saying and why, you're in good shape for quiz and exam questions.  Type Description Jump Each of the one-sided limits are finite but disagree Removable The two one-sided limits agree but the function value does not agree Infinite One or both of the one-sided limits is either$+\infty$or$-\infty$Oscillating Typically examining some form of$\sin(\theta)$or$\cos(\theta)$at points where$\theta=1/0^+$or$\theta=1/0^-$## One-Sided Continuity Just like limits can be one-sided, so too can continuity. Knowing and understanding this nuance is less important than understanding everything we've mentioned up to this point, and many teachers will not even discuss one-sided continuity, so if you don't need to know about it, jump down to the next heading. If you do, read on!Jump discontinuities may be considered continuous on one side by a very similar definition to the general limit definition of continuity, but using one-sided limits: Definition:A function$f(x)$is continuous from the left if$$\lim_{x \to a^-} f(x) = f(a)$$and continuous from the right if$$\lim_{x \to a^+} f(x) = f(a)$$ Said a different way, jump discontinuities are one-sidedly continuous if one side has the filled in solid dot, and not one-sidedly continuous if the function is defined elsewhere.  Note that this will only be possible for jump discontinuities. ## Well-Behaved Functions By now, you may have noticed that many of the continuity problems we are describing are specific to specialized functions, like piecewise, rational, or trigonometric functions. This is no coincidence. Many common functions, such as polynomials, have no instances of discontinuity at any point. In fact, several of the major function families have continuity behavior that can be summarized generally. Theorem 1:The following function families are continuous at every point in their domain: • Linear and Constant Functions • Polynomial Functions • Rational Functions • Exponential Functions • Logarithmic Functions • Trigonometric Functions But wait, didn't we just say that$f(x) = \sin(1/x)$had that strange infinite discontinuity at$x=0$? We sure did, but that's still in line with this theorem -$x=0$is not in the domain of$\sin(1/x)$, and so the theorem does not guarantee continuity for this function at$x=0$. It's noteworthy that the function promises continuity for every other real number$x$that could be input into$\sin(1/x)$, since$x=0$is the only number excluded from the domain.The takeaway is that, while it is possible to design funky and unusual variations of these typical function families, we can always guarantee continuity of these function types for any value of$x$that could be input into it to obtain a real answer (i.e. the domain). ## Mr. Math Makes It Mean The last consideration about continuity that we need to make is the case where some function$h(x)$is defined as some combination of two (or possibly more) other functions, say$f(x)$and$g(x)$. This could happen by adding, subtracting, multiplying, or dividing$f$and$g$, or it could be a result of composition, e.g.$h(x) = f(g(x))$. In any case, there is nothing too surprising happening here. Definition:Let$h(x)$be comprised of two functions,$f(x)$and$g(x)$, either by arithmetic or composition. It follows that$h(x)$will be continuous everywhere that$f(x)$and$g(x)$are both continuous. For function division, such as$h(x) = f(x) \, / \, g(x)$for example, we must also exclude$x$values that make$g(x)=0$, for$h(x) = f(x) \, / \, g(x)$would then be undefined. Similarly, we must exclude any$x$values that make$g(x)$equal to values that are not in the domain of$f(x)$for a composition of the form$f\big( g(x) \big)$, as to avoid causing domain issues with$f$. It is not a surprising fact that domain and continuity go hand-in-hand, but it's worth mentioning explicitly for clarity, and teachers will occasionally build quiz questions around this idea. ## Put It To The Test Identify Continuity - Graph BasedAmong the warm-up questions we should expect are questions asking us to identify which continuity type is which, based on the graph:Instructions for Examples 1-3: Determine the intervals on which the function is continuous, and determine the category of discontinuity that occurs anywhere that the function is discontinuous. Example 1 Show solution$\blacktriangleright$In this graph, not only must we identify the discontinuities at$x=-4$and$x=4$, but we must also be aware that the function doesn't appear to be defined at all between$x=2$and$x=4$. This is a good reminder that when we report the intervals of continuity, that we must not blindly expect the answer to be all real numbers excepting the discontinuity points. The answer is that the function is continuous on the following intervals:$$(-\infty, -4) \,\, \bigcup \,\, (-4, 2) \,\, \bigcup \,\, (4, \infty)$$Additionally, discontinuities exist at$x=-4$(of the jump variety) and$x=4$(of the infinite variety). Example 2 Show solution$\blacktriangleright$This function has two discontinuities - one at$-1$and one at$5$. The point labeled at$3$has no discontinuity - it was randomly labeled. Some teachers do that, so don't expect every labeled point to have a discontinuity issue. The discontinuity at$x=-1$is a jump type with right-side continuity, and the discontinuity at$x=3$is a removable discontinuity. Therefore we would answer the question by giving the union of the intervals of continuity:$$(-\infty, -1) \,\, \bigcup \,\, (-1, 5) \,\, \bigcup \,\, (5, \infty)$$Note that we do not notate the one-sided continuity unless we're asked about it, as including the point$-1$in the answer would suggest that the function is continuous at$-1$. Example 3 Show solution$\blacktriangleright$While you'd be correct to say this function isn't "smooth" (more to come on that soon, in the lesson that discusses differentiability »), it is not discontinuous at the corner point. In fact, this function looks to be continuous everywhere. Therefore, we will report that the function is continuous on the interval$$(-\infty, \infty)$$And because there are no discontinuities, we don't have to state types of discontinuities, or label anything further. Identify Continuity - Symbol BasedSimilarly yet more algebraically involved, we are often asked to classify based on function definition alone.Instructions for Examples 4-6: Determine the intervals on which the function is continuous, and determine the category of discontinuity that occurs anywhere that the function is discontinuous. Example 4$$f(x) = \frac{3x}{x^2-9}$$ Show solution$\blacktriangleright$Since rational functions are continuous everywhere in their domain (via Theorem 1 of this lesson), we only need to identify points that are not in the domain of$f(x)$when seeking points of potential discontinuity. In this case, the function is undefined when the denominator is zero, which happens at$x=-3$and$x=3$. Via our limit evaluation skills, you should conclude that$$\lim_{x \to -3^-} = -\infty$$$$\lim_{x \to -3^+} = \infty$$$$\lim_{x \to 3^-} = -\infty$$$$\lim_{x \to 3^+} = \infty$$Therefore we know that$f(x)$is continuous everywhere except for at$x=-3$and$x=3$, and that each of these discontinuities is of the infinite category. Example 5$$f(x) = \frac{2x^2 - 7x + 4}{3x^2 - 5x - 12}$$ Show solution$\blacktriangleright$Once again, faced with a rational function, we should seek to identify$x$values that are not in the domain of the function, since a rational function such as this one will be continuous at all other$x$values beside ones that are not in the domain (again, per Theorem 1 above). In the absence of logs, radicals, or anything funky, this rational function's domain will be all real numbers except for places that make the denominator zero.When is$3x^2 - 5x - 12 = 0$? Let's factor it and find out.$$3x^2 - 5x - 12 = (x-3)(3x+4) = 0$$$$\therefore x=3, \,\,\,\, x = \frac{-4}{3}$$Note that if we couldn't factor this denominator, that would be ok - we would just go straight to using the quadratic formula in that scenario.By Theorem 1, we conclude that there is a discontinuity in two places: at$x=3$and$x = -4/3$. Now we just need to determine what category of discontinuity each point falls into.If we factor both the numerator and the denominator, we will see that$$\frac{2x^2 - 7x + 4}{3x^2 - 5x - 12} = \frac{(x-3)(2x-1)}{(x-3)(3x+4)}$$$$= \frac{2x-1}{3x+4}$$From the knowledge we picked up when studying limits with zero in the denominator », this tells us that 1)$x=3$is a puncture, not an asymptote, and 2)$x=-4/3$is a vertical asymptote. Therefore, we can conclude that:$f(x)$has a removable discontinuity at$x=3$$f(x) has an infinite discontinuity at x = -4/3 Warning! While limit evaluation techniques dictate that we should seek to factor and cancel in a rational function to more quickly identify limit results, be careful not to jump to this process right away for questions regarding continuity. In Example 5, if you were to factor and cancel immediately, you would miss one of the places that is not in the domain of the function, because you removed the factor from the picture completely. Just because a denominator factor cancels with a numerator factor does not mean the associated root is in the domain of the original function - here x=3 is not in the domain of f(x), even though the (x-3) factors cancel, and it would be incorrect to ignore the discontinuity at that point. Example 6$$ f(x) = \left\{ \begin{array}{ll} -2x + 1 & : x \le -1\\ x^2 - 3 & : -1 \gt x \gt 3\\ \sqrt{x-3} & : x \ge 3 \end{array} \right. $$Show solution \blacktriangleright On their own, each of these three function definitions is continuous everywhere on their domain (again, via Theorem 1), so the only potential places we need to investigate are the breakpoints of the piecewise function.Let's start by examining the behavior at x=-1:$$\lim_{x \to -1^{-}} f(x) = \lim_{x \to -1^{-}} -2x + 1 = 3\lim_{x \to -1^{+}} f(x) = \lim_{x \to -1^{+}} x^2 - 3 = -2$$Since each one sided limit yields different but finite values, we know immediately that there is a jump discontinuity at x=-1. Now let's look at x = 3:$$\lim_{x \to 3^{-}} f(x) = \lim_{x \to 3^{-}} x^2 - 3 = 6\lim_{x \to 3^{+}} f(x) = \lim_{x \to 3^{+}} \sqrt{x-3} = 0$$Once again, we have a jump discontinuity. Therefore, we conclude that f(x) is continuous everywhere except at x=-1 and x=3, where in each case there exists a jump discontinuity.Combinations of FunctionsIn examples 7-9, use the given information and your knowledge of function combination to answer each question. Example 7The domain of f(x) is all real numbers, and it is continuous everywhere on its domain.The domain of g(x) is all real numbers except x=-2 and x=2, and is continuous everywhere on its domain.Find the intervals of continuity for the function (f + 2g)(x). Show solution \blacktriangleright Note that multiplying g(x) by 2 doesn't affect its domain - it does affect its range, but we will not need to consider range to answer this question.The given information about f and g implies that the combined function (f + 2g)(x) would have a domain of all real numbers except for x=-2 and x=2. And since both functions are continuous everywhere on their domain, it follows that the combined function is as well. Therefore, we can safely conclude that the combined function is continuous on the domain$$(-\infty, -2) \,\, \bigcup \,\, (-2, 2) \,\, \bigcup \,\, (2, \infty)$$Example 8The domain of f(x) is all real numbers except x=1, and f is continuous everywhere on its domain.The domain of g(x) is all real numbers except x=4, and g is continuous everywhere on its domain.Find the intervals of continuity for the function \bigg(\frac{f}{g}\bigg)(x). Show solution \blacktriangleright Similar to the last problem, the key is to follow the logic that governs the domain of a combination function, and let the continuity question follow the same logic. The domain of a quotient of functions is found by including only numbers that work in both functions simultaneously, which in this case would be all real numbers except x=1 and x=4. However, we also need to ensure that we do not create a division by zero error, and so we must also exclude any x that makes g(x) equal to zero, as this would cause f(x) / g(x) to be undefined. Since we do not have the functions written for us explicitly, the best way to answer the question is to say the following:The domain of \bigg(\frac{f}{g}\bigg)(x) is all real numbers except x=1, x=4, and any value of x such that g(x)=0. Because both functions are continuous at every point on their domain, we conclude that the function \bigg(\frac{f}{g}\bigg)(x) is continuous at all values of x except for x=1, x=4, and any value of x such that g(x)=0. Example 9The domain of g(x) is all real numbers except x=a, and f is continuous everywhere on its domain.The domain of h(x) is all real numbers except x=b, and g is continuous everywhere on its domain.a and b are real number constants.Find the intervals of continuity for the function  g \big( h(x) \big). Show solution \blacktriangleright As with the last two problems, the domain is the key to making a statement about where the composite function is and is not continuous. The domain of  g \big( h(x) \big) will be all real numbers except for x=a, x=b, and any x value that makes h(x)=a, since a cannot be input into the function g(x). Therefore, we will report our answer asThe function  g \big( h(x) \big) is continuous at all values of x except for x=a, x=b, and any x value such that h(x)=a. Unknown Coefficient Problems Finally, we may be given piecewise functions with unknown coefficients. We can identify what coefficient values must be present to make the function continuous:Instructions for Examples 10-11: Determine the unknown coefficients a, b, etc such that the function is continuous everywhere on its domain.Example 10$$ f(x) = \left\{ \begin{array}{lr} \cos(x) & : x \le 0\\ \sin(x) + a & : x \gt 0 \end{array} \right. $$Show solution \blacktriangleright As we've seen with common functions that are covered by Theorem 1, we only need to worry about the piecewise function break point. Let's examine the limits from each side as x approaches 0.$$\lim_{x \to 0^-} \cos(x) = 1\lim_{x \to 0^+} \sin(x) + a = 0 + a = a$$All we need to do is make each one-sided limit equal to one another. There isn't too much to do in this particular example, so we have1 = aand therefore, the function will be continuous if a=1. Example 11$$ f(x) = \left\{ \begin{array}{lr} 7 & : x < -2\\ x^2 - 2ax + b & : -2 \leq x < 6\\ ax + b & : x \geq 6 \end{array} \right. $$Show solution \blacktriangleright As with the last example, we seek to find the constants that make each piece of the piecewise function equal to one another at the interval break points. Let's work left-to-right and see what we can discern: set the left-most function definition equal to the middle function definition.$$7 = x^2 - 2ax + b$$Furthermore, we want these two parts of the function definition to be equivalent specifically at x=-2, where the piecewise function "changes over" from the first definition to the second. So let's let x be -2.$$7 = 4 + 4a + b$$This equation is indeterminate, having 2 variables in it. Let's pause on it and do the same thing for the other cutoff point of the piecewise function - set the two pieces equal.$$x^2 - 2ax + b = ax + b$$At the place that we want these two definitions to be equal, x is 6. Let's plug in that value and see what remains.$$36 - 12a + \cancel{b} = 6a + \cancel{b}36 - 18a = 0\boxed{a = 2}$$Now that we know a=2, we can use that result in the first equation we had set up but had to put on pause:$$7 = 4 + 4a + b7 = 4 + 4(2) + b\boxed{b = -5}$$Therefore, b = -5, and the complete piecewise function that is continuous at all points is$$ f(x) = \left\{ \begin{array}{lr} 7 & : x < -2\\ x^2 - 4x - 5 & : -2 \leq x < 6\\ 2x -5 & : x \geq 6 \end{array} \right. If we wanted to double check, we can plug in this function into a graphing calculator to see that the various piecewise parts do indeed connect at the "break points". You Should Know Problems like these are not only intended to be worked without a graph, but may actually become incredibly confusing should you to try to graph them. Their true picture is not known at the start of the problem, since the graph would depend on the unknown constants that you are solving for.

Lesson Takeaways
• Understand what continuity is visually, conceptually, and symbolically
• Be able to classify discontinuity types based on graphs
• Be able to classify discontinuity types without graphs, by analyzing with limits
• Concisely describe the places or intervals on which a function is continuous
• If required, be able to identify one-sided continuity
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