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Continuity

Determine whether a function is continuous at a number.

Continuity at a Point

Before we look at a formal definition of what it means for a function 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 . 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:

\[\text{i.}\ f(a)\ \text{is defined.}\]

However, as we see in , 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 $\underset{x\to a}{\text{lim}}f(x)$ does not exist. We must add another condition for continuity at a—namely,

\[\text{ii.}\ \underset{x\to a}{\text{lim}}f(x)\ \text{exists.}\]

However, as we see in , 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:

\[\text{iii.}\ \underset{x\to a}{\text{lim}}f(x)=f(a).\]

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

The following procedure can be used to analyze the continuity of a function at a point using this definition.

Determining Continuity at a Point

Check to see if $f(a)$ is defined. If $f(a)$ is undefined, we need go no further. The function is not continuous at a. If $f(a)$ is defined, continue to step 2.
Compute $\underset{x\to a}{\text{lim}}f(x).$ In some cases, we may need to do this by first computing $\underset{x\to {a}^{-}}{\text{lim}}f(x)$ and $\underset{x\to {a}^{+}}{\text{lim}}f(x).$ If $\underset{x\to a}{\text{lim}}f(x)$ does not exist (that is, it is not a real number), then the function is not continuous at a and the problem is solved. If $\underset{x\to a}{\text{lim}}f(x)$ exists, then continue to step 3.
Compare $f(a)$ and $\underset{x\to a}{\text{lim}}f(x).$ If $\underset{x\to a}{\text{lim}}f(x)\ne f(a),$ then the function is not continuous at a. If $\underset{x\to a}{\text{lim}}f(x)=f(a),$ then the function is continuous at a.

Condensed — the full section is in OpenStax Calculus Volume 1.

Types of Discontinuities

As we have seen in and , discontinuities take on several different appearances. We classify the types of discontinuities we have seen thus far as removable discontinuities, infinite discontinuities, or jump discontinuities. Intuitively, a removable discontinuity is a discontinuity for which there is a hole in the graph, a jump discontinuity is a noninfinite discontinuity for which the sections of the function do not meet up, and an infinite discontinuity is a discontinuity located at a vertical asymptote. illustrates the differences in these types of discontinuities. Although these terms provide a handy way of describing three common types of discontinuities, keep in mind that not all discontinuities fit neatly into these categories.

These three discontinuities are formally defined as follows:

Definition

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

$f$ has a removable discontinuity at a if $\underset{x\to a}{\text{lim}}f(x)$ exists. (Note: When we state that $\underset{x\to a}{\text{lim}}f(x)$ exists, we mean that $\underset{x\to a}{\text{lim}}f(x)=L,$ where L is a real number.)
$f$ has a jump discontinuity at a if $\underset{x\to {a}^{-}}{\text{lim}}f(x)$ and $\underset{x\to {a}^{+}}{\text{lim}}f(x)$ both exist, but $\underset{x\to {a}^{-}}{\text{lim}}f(x)\ne \underset{x\to {a}^{+}}{\text{lim}}f(x).$ (Note: When we state that $\underset{x\to {a}^{-}}{\text{lim}}f(x)$ and $\underset{x\to {a}^{+}}{\text{lim}}f(x)$ both exist, we mean that both are real-valued and that neither take on the values ±∞.)
$f$ has an infinite discontinuity at a if $\underset{x\to {a}^{-}}{\text{lim}}f(x)=\text{\pm }\infty$ and/or $\underset{x\to {a}^{+}}{\text{lim}}f(x)=\text{\pm }\infty .$

Example

Try it.

In , 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 $\underset{x\to 2}{\text{lim}}f(x)\text{:}$

\[\begin{array}{ll}\underset{x\to 2}{\text{lim}}f(x) & =\underset{x\to 2}{\text{lim}}\frac{{x}^{2}-4}{x-2} \\ & =\underset{x\to 2}{\text{lim}}\frac{(x-2)(x+2)}{x-2} \\ & =\underset{x\to 2}{\text{lim}}(x+2) \\ & =4.\end{array}\]

Since f is discontinuous at 2 and $\underset{x\to 2}{\text{lim}}f(x)$ exists, f has a removable discontinuity at $x=2.$

Condensed — the full section is in OpenStax Calculus Volume 1.

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 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 looking at the definition of what it means for a function to be continuous from the right at a point and continuous from the left at a point.

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 are defined in a similar fashion.

Requiring that $\underset{x\to {a}^{+}}{\text{lim}}f(x)=f(a)$ and $\underset{x\to {b}^{-}}{\text{lim}}f(x)=f(b)$ 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, $\underset{x\to {a}^{+}}{\text{lim}}f(x)\ne f(a),$ 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

Try it.

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 $(\text{-}\infty ,-2)\cup (-2,0)\cup (0,\text{+}\infty ).$ Thus, $f(x)$ is continuous over each of the intervals $(\text{-}\infty ,-2),(-2,0),$ and $(0,\text{+}\infty ).$

The 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.

Condensed — the full section is in OpenStax Calculus Volume 1.

Proof

We begin by demonstrating that $\text{cos}\ x$ is continuous at every real number. To do this, we must show that $\underset{x\to a}{\text{lim}}\text{cos}\ x=\text{cos}\ a$ for all values of a.

$\begin{array}{lllll}\underset{x\to a}{\text{lim}}\text{cos}\ x & =\underset{x\to a}{\text{lim}}\text{cos}\ ((x-a)+a) & & & \text{rewrite}\ x=x-a+a \\ & =\underset{x\to a}{\text{lim}}(\text{cos}\ (x-a)\ \text{cos}\ a-\text{sin}\ (x-a)\ \text{sin}\ a) & & & \text{apply the identity for the cosine of the sum of two angles} \\ & =\text{cos}\ (\underset{x\to a}{\text{lim}}(x-a))\ \text{cos}\ a-\text{sin}\ (\underset{x\to a}{\text{lim}}(x-a))\ \text{sin}\ a & & & \underset{x\to a}{\text{lim}}(x-a)=0,\ \text{and}\ \text{sin}\ x\ \text{and}\ \text{cos}\ x\ \text{are continuous at 0} \\ & =\text{cos}\ (0)\ \text{cos}\ a-\text{sin}\ (0)\ \text{sin}\ a & & & \text{evaluate cos(0) and sin(0) and simplify} \\ & =1\cdot \text{cos}\ a-0\cdot \text{sin}\ a=\text{cos}\ a.\end{array}$

The proof that $\text{sin}\ x$ is continuous at every real number is analogous. Because the remaining trigonometric functions may be expressed in terms of $\text{sin}\ x$ and $\text{cos}\ x,$ their continuity follows from the quotient limit law.

□

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.

Example

Try it.

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

Solution

Since $f(x)=x-\text{cos}\ x$ is continuous over $(\text{-}\infty ,\text{+}\infty ),$ 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

\[f(0)=0-\text{cos}\ (0)=-1<0\]

and

\[f(\frac{\pi }{2})=\frac{\pi }{2}-\text{cos}\frac{\pi }{2}=\frac{\pi }{2}>0.\]

Using the Intermediate Value Theorem, we can see that there must be a real number c in $[0,\pi \text{/}2]$ that satisfies $f(c)=0.$ Therefore, $f(x)=x-\text{cos}\ x$ has at least one zero.

Example

Try it.

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 $[0,2\text{]?}$ 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

Try it.

For $f(x)=1\text{/}x,f(-1)=-1<0$ 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.

Key Concepts

For a function to be continuous at a point, it must be defined at that point, its limit must exist at the point, and the value of the function at that point must equal the value of the limit at that point.
Discontinuities may be classified as removable, jump, or infinite.
A function is continuous over an open interval if it is continuous at every point in the interval. It is continuous over a closed interval if it is continuous at every point in its interior and is continuous at its endpoints.
The composite function theorem states: If $f(x)$ is continuous at L and $\underset{x\to a}{\text{lim}}g(x)=L,$ then $\underset{x\to a}{\text{lim}}f(g(x))=f(\underset{x\to a}{\text{lim}}g(x))=f(L).$
The Intermediate Value Theorem guarantees that if a function is continuous over a closed interval, then the function takes on every value between the values at its endpoints.

Continuity

For the following exercises, determine the point(s), if any, at which each function is discontinuous. Classify any discontinuity as jump, removable, infinite, or other.

For the following exercises, decide if the function continuous at the given point. If it is discontinuous, what type of discontinuity is it?

In the following exercises, find the value(s) of k that makes each function continuous over the given interval.

In the following exercises, use the Intermediate Value Theorem (IVT).

In the following exercises, suppose $y=f(x)$ is defined for all x. For each description, sketch a graph with the indicated property.

Determine whether each of the given statements is true. Justify your response with an explanation or counterexample.

[T] The following problems consider the scalar form of Coulomb’s law, which describes the electrostatic force between two point charges, such as electrons. It is given by the equation $F(r)={k}_{e}\frac{|{q}_{1}{q}_{2}|}{{r}^{2}},$ where ${k}_{e}$ is Coulomb’s constant, ${q}_{i}$ are the magnitudes of the charges of the two particles, and r is the distance between the two particles.

Condensed — the full section is in OpenStax Calculus Volume 1.

Determining Whether a Function Is Continuous at a Number

Let’s consider a specific example of temperature in terms of date and location, such as June 27, 2013, in Phoenix, AZ. The graph in indicates that, at 2 a.m., the temperature was ${96}^{∘}\text{F}$. By 2 p.m. the temperature had risen to ${116}^{∘}\text{F,}$ and by 4 p.m. it was ${118}^{∘}\text{F}\text{.}$ Sometime between 2 a.m. and 4 p.m., the temperature outside must have been exactly ${110.5}^{∘}\text{F}\text{.}$ In fact, any temperature between ${96}^{∘}\text{F}$ and ${118}^{∘}\text{F}$ occurred at some point that day. This means all real numbers in the output between ${96}^{∘}\text{F}$ and ${118}^{∘}\text{F}$ are generated at some point by the function according to the intermediate value theorem,

Look again at . There are no breaks in the function’s graph for this 24-hour period. At no point did the temperature cease to exist, nor was there a point at which the temperature jumped instantaneously by several degrees. A function that has no holes or breaks in its graph is known as a continuous function. Temperature as a function of time is an example of a continuous function.

If temperature represents a continuous function, what kind of function would not be continuous? Consider an example of dollars expressed as a function of hours of parking. Let’s create the function $D,$ where $D(x)$ is the output representing cost in dollars for parking $x$ number of hours. See .

Suppose a parking garage charges $4.00 per hour or fraction of an hour, with a $25 per day maximum charge. Park for two hours and five minutes and the charge is $12. Park an additional hour and the charge is $16. We can never be charged $13, $14, or $15. There are real numbers between 12 and 16 that the function never outputs. There are breaks in the function’s graph for this 24-hour period, points at which the price of parking jumps instantaneously by several dollars.

A function that remains level for an interval and then jumps instantaneously to a higher value is called a stepwise function. This function is an example.

A function that has any hole or break in its graph is known as a discontinuous function. A stepwise function, such as parking-garage charges as a function of hours parked, is an example of a discontinuous function.

So how can we decide if a function is continuous at a particular number? We can check three different conditions. Let’s use the function $y=f(x)$ represented in as an example.

Condensed — the full section is in OpenStax Precalculus 2e.

Recognizing Continuous and Discontinuous Real-Number Functions

Many of the functions we have encountered in earlier chapters are continuous everywhere. They never have a hole in them, and they never jump from one value to the next. For all of these functions, the limit of $f(x)$ as $x$ approaches $a$ is the same as the value of $f(x)$ when $x=a.$ So $\underset{x\to a}{\lim }f(x)=f(a).$ There are some functions that are continuous everywhere and some that are only continuous where they are defined on their domain because they are not defined for all real numbers.

Example

Try it.

Determine whether the function $f(x)=\frac{{x}^{2}-25}{x-5}$ is continuous at $x=5.$

Solution

To determine if the function $f$ is continuous at $x=5,$ we will determine if the three conditions of continuity are satisfied at $x=5.$

Condition 1:

\[\begin{array}{l}f(5)\text{ does not exist}\text{.} \\ ⇒\text{Condition 1 fails}.\end{array}\]

There is no need to proceed further. Condition 2 fails at $x=5.$ If any of the conditions of continuity are not satisfied at $x=5,$ the function $f$ is not continuous at $x=5.$

Condensed — the full section is in OpenStax Precalculus 2e.

Determining the Input Values for Which a Function Is Discontinuous

Now that we can identify continuous functions, jump discontinuities, and removable discontinuities, we will look at more complex functions to find discontinuities. Here, we will analyze a piecewise function to determine if any real numbers exist where the function is not continuous. A piecewise function may have discontinuities at the boundary points of the function as well as within the functions that make it up.

To determine the real numbers for which a piecewise function composed of polynomial functions is not continuous, recall that polynomial functions themselves are continuous on the set of real numbers. Any discontinuity would be at the boundary points. So we need to explore the three conditions of continuity at the boundary points of the piecewise function.

Condensed — the full section is in OpenStax Precalculus 2e.

Determining Whether a Function Is Continuous

To determine whether a piecewise function is continuous or discontinuous, in addition to checking the boundary points, we must also check whether each of the functions that make up the piecewise function is continuous.

Example

Try it.

Determine whether the function below is continuous. If it is not, state the location and type of each discontinuity.

\[f(x)=\{\begin{array}{ll}\sin (x), & x<0 \\ {x}^{3}, & x>0\end{array}\]

Solution

The two functions composing this piecewise function are $f(x)=\sin (x)$ on $x<0$ and $f(x)={x}^{3}$ on $x>0.$ The sine function and all polynomial functions are continuous everywhere. Any discontinuities would be at the boundary point,

At $x=0,$ let us check the three conditions of continuity.

Condition 1:

\[\begin{array}{l}f(0)\text{ does not exist}. \\ ⇒\text{Condition 1 fails}.\end{array}\]

Because all three conditions are not satisfied at $x=0,$ the function $f(x)$ is discontinuous at $x=0.$

Key Concepts

A continuous function can be represented by a graph without holes or breaks.
A function whose graph has holes is a discontinuous function.
A function is continuous at a particular number if three conditions are met:
- Condition 1: $f(a)$ exists.
- Condition 2: $\underset{x\to a}{\lim }f(x)$ exists at $x=a.$
- Condition 3: $\underset{x\to a}{\lim }f(x)=f(a).$
A function has a jump discontinuity if the left- and right-hand limits are different, causing the graph to “jump.”
A function has a removable discontinuity if it can be redefined at its discontinuous point to make it continuous. See .
Some functions, such as polynomial functions, are continuous everywhere. Other functions, such as logarithmic functions, are continuous on their domain. See and .
For a piecewise function to be continuous each piece must be continuous on its part of the domain and the function as a whole must be continuous at the boundaries. See and .

Practice (40)

Try each one on paper first. Reveal the answer to check; verified ones can be opened in the solver for every step.

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

ჲრკპთირვ ჲრდჲგჲპა.

Let’s begin by trying to calculate $f(2).$ We can see that $f(2)=0\text{/}0,$ 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 .
Using the definition, determine whether the function $f(x)=\{\begin{array}{ll}-{x}^{2}+4 & \text{if}\ x\le 3 \\ 4x-8 & \text{if}\ x>3\end{array}$ is continuous at $x=3.$ Justify the conclusion.

ჲრკპთირვ ჲრდჲგჲპა.

Let’s begin by trying to calculate $f(3).$
\[f(3)=-({3}^{2})+4=-5.\]
Thus, $f(3)$ is defined. Next, we calculate $\underset{x\to 3}{\text{lim}}f(x).$ To do this, we must compute $\underset{x\to {3}^{-}}{\text{lim}}f(x)$ and $\underset{x\to {3}^{+}}{\text{lim}}f(x)\text{:}$
\[\underset{x\to {3}^{-}}{\text{lim}}f(x)=-({3}^{2})+4=-5\]
and
\[\underset{x\to {3}^{+}}{\text{lim}}f(x)=4(3)-8=4.\]
Therefore, $\underset{x\to 3}{\text{lim}}f(x)$ does not exist. Thus, $f(x)$ is not continuous at 3. The graph of $f(x)$ is shown in .
Using the definition, determine whether the function $f(x)=\{\begin{array}{ll}\frac{\text{sin}\ x}{x} & \text{if}\ x\ne 0 \\ 1 & \text{if}\ x=0\end{array}$ is continuous at $x=0.$

ჲრკპთირვ ჲრდჲგჲპა.

First, observe that
\[f(0)=1.\]
Next,
\[\underset{x\to 0}{\text{lim}}f(x)=\underset{x\to 0}{\text{lim}}\frac{\text{sin}\ x}{x}=1.\]
Last, compare $f(0)$ and $\underset{x\to 0}{\text{lim}}f(x).$ We see that
\[f(0)=1=\underset{x\to 0}{\text{lim}}f(x).\]
Since all three of the conditions in the definition of continuity are satisfied, $f(x)$ is continuous at $x=0.$
Using the definition, determine whether the function $f(x)=\{\begin{array}{ll}2x+1 & \text{if}\ x<1 \\ 2 & \text{if}\ x=1 \\ -x+4 & \text{if}\ x>1\end{array}$ 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.

ჲრკპთირვ ჲრდჲგჲპა.

f is not continuous at 1 because $f(1)=2\ne 3=\underset{x\to 1}{\text{lim}}f(x).$
For what values of x is $f(x)=\frac{x+1}{x-5}$ continuous?

ჲრკპთირვ ჲრდჲგჲპა.

The rational function $f(x)=\frac{x+1}{x-5}$ is continuous for every value of x except $x=5.$
For what values of x is $f(x)=3{x}^{4}-4{x}^{2}$ continuous?

ჲრკპთირვ ჲრდჲგჲპა.

$f(x)$ is continuous at every real number.
In , we showed that $f(x)=\frac{{x}^{2}-4}{x-2}$ is discontinuous at $x=2.$ Classify this discontinuity as removable, jump, or infinite.

ჲრკპთირვ ჲრდჲგჲპა.

To classify the discontinuity at 2 we must evaluate $\underset{x\to 2}{\text{lim}}f(x)\text{:}$
\[\begin{array}{ll}\underset{x\to 2}{\text{lim}}f(x) & =\underset{x\to 2}{\text{lim}}\frac{{x}^{2}-4}{x-2} \\ & =\underset{x\to 2}{\text{lim}}\frac{(x-2)(x+2)}{x-2} \\ & =\underset{x\to 2}{\text{lim}}(x+2) \\ & =4.\end{array}\]
Since f is discontinuous at 2 and $\underset{x\to 2}{\text{lim}}f(x)$ exists, f has a removable discontinuity at $x=2.$
In , we showed that $f(x)=\{\begin{array}{ll}-{x}^{2}+4 & \text{if}\ x\le 3 \\ 4x-8 & \text{if}\ x>3\end{array}$ is discontinuous at $x=3.$ Classify this discontinuity as removable, jump, or infinite.

ჲრკპთირვ ჲრდჲგჲპა.

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

ჲრკპთირვ ჲრდჲგჲპა.

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 $\underset{x\to {-1}^{-}}{\text{lim}}\frac{x+2}{x+1}=\text{-}\infty$ and $\underset{x\to {-1}^{+}}{\text{lim}}\frac{x+2}{x+1}=\text{+}\infty .$ Therefore, the function has an infinite discontinuity at −1.
For $f(x)=\{\begin{array}{ll}{x}^{2} & \text{if}\ x\ne 1 \\ 3 & \text{if}\ x=1\end{array},$ decide whether f is continuous at 1. If f is not continuous at 1, classify the discontinuity as removable, jump, or infinite.

ჲრკპთირვ ჲრდჲგჲპა.

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

ჲრკპთირვ ჲრდჲგჲპა.

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 $(\text{-}\infty ,-2)\cup (-2,0)\cup (0,\text{+}\infty ).$ Thus, $f(x)$ is continuous over each of the intervals $(\text{-}\infty ,-2),(-2,0),$ and $(0,\text{+}\infty ).$
State the interval(s) over which the function $f(x)=\sqrt{4-{x}^{2}}$ is continuous.

ჲრკპთირვ ჲრდჲგჲპა.

From the limit laws, we know that $\underset{x\to a}{\text{lim}}\sqrt[]{4-{x}^{2}}=\sqrt{4-{a}^{2}}$ for all values of a in $(-2,2).$ We also know that $\underset{x\to {-2}^{+}}{\text{lim}}\sqrt{4-{x}^{2}}=0$ exists and $\underset{x\to {2}^{-}}{\text{lim}}\sqrt{4-{x}^{2}}=0$ exists. Therefore, $f(x)$ is continuous over the interval $[-2,2].$
State the interval(s) over which the function $f(x)=\sqrt{x+3}$ is continuous.

ჲრკპთირვ ჲრდჲგჲპა.

$[-3,\text{+}\infty )$
Evaluate $\underset{x\to \pi \text{/}2}{\text{lim}}\text{cos}\ (x-\frac{\pi }{2}).$

ჲრკპთირვ ჲრდჲგჲპა.

The given function is a composite of $\text{cos}\ x$ and $x-\frac{\pi }{2}.$ Since $\underset{x\to \pi \text{/}2}{\text{lim}}(x-\frac{\pi }{2})=0$ and $\text{cos}\ x$ is continuous at 0, we may apply the composite function theorem. Thus,
\[\underset{x\to \pi \text{/}2}{\text{lim}}\text{cos}\ (x-\frac{\pi }{2})=\text{cos}\ (\underset{x\to \pi \text{/}2}{\text{lim}}(x-\frac{\pi }{2}))=\text{cos}\ (0)=1.\]
Evaluate $\underset{x\to \pi }{\text{lim}}\text{sin}\ (x-\pi ).$

ჲრკპთირვ ჲრდჲგჲპა.

0
Show that $f(x)=x-\text{cos}\ x$ has at least one zero.

ჲრკპთირვ ჲრდჲგჲპა.

Since $f(x)=x-\text{cos}\ x$ is continuous over $(\text{-}\infty ,\text{+}\infty ),$ 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
\[f(0)=0-\text{cos}\ (0)=-1<0\]
and
\[f(\frac{\pi }{2})=\frac{\pi }{2}-\text{cos}\frac{\pi }{2}=\frac{\pi }{2}>0.\]
Using the Intermediate Value Theorem, we can see that there must be a real number c in $[0,\pi \text{/}2]$ that satisfies $f(c)=0.$ Therefore, $f(x)=x-\text{cos}\ x$ has at least one zero.
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 $[0,2\text{]?}$ Explain.

ჲრკპთირვ ჲრდჲგჲპა.

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

ჲრკპთირვ ჲრდჲგჲპა.

No. The function is not continuous over $[-1,1].$ The Intermediate Value Theorem does not apply here.
Show that $f(x)={x}^{3}-{x}^{2}-3x+1$ has a zero over the interval $[0,1].$

ჲრკპთირვ ჲრდჲგჲპა.

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

ჲრკპთირვ ჲრდჲგჲპა.

The function is defined for all x in the interval $(0,\infty ).$
$f(x)=\frac{2}{{x}^{2}+1}$
$f(x)=\frac{x}{{x}^{2}-x}$

ჲრკპთირვ ჲრდჲგჲპა.

Removable discontinuity at $x=0;$ infinite discontinuity at $x=1$
$g(t)={t}^{-1}+1$
$f(x)=\frac{5}{{e}^{x}-2}$

ჲრკპთირვ ჲრდჲგჲპა.

Infinite discontinuity at $x=\text{ln}\ 2$
$f(x)=\frac{|x-2|}{x-2}$
$H(x)=\text{tan}\ 2x$

ჲრკპთირვ ჲრდჲგჲპა.

Infinite discontinuities at $x=\frac{(2k+1)\pi }{4},$ for $k=0,\pm 1,\pm 2,\pm 3\text{,\ldots }$
$f(t)=\frac{t+3}{{t}^{2}+5t+6}$
$f(x)=\frac{2{x}^{2}-5x+3}{x-1}$ at $x=1$

ჲრკპთირვ ჲრდჲგჲპა.

No. It is a removable discontinuity.
$h(\theta )=\frac{\text{sin}\ \theta -\text{cos}\ \theta }{\text{tan}\ \theta }$ at $\theta =\pi$
$g(u)=\{\begin{array}{ll}\frac{6{u}^{2}+u-2}{2u-1} & \text{if}\ u\ne \frac{1}{2} \\ \frac{7}{2} & \text{if}\ u=\frac{1}{2}\end{array},$ at $u=\frac{1}{2}$

ჲრკპთირვ ჲრდჲგჲპა.

Yes. It is continuous.
$f(y)=\frac{\text{sin}\ (\pi y)}{\text{tan}\ (\pi y)},$ at $y=1$
$f(x)=\{\begin{array}{ll}{x}^{2}-{e}^{x} & \text{if}\ x<0 \\ x-1 & \text{if}\ x\ge 0\end{array},$ at $x=0$

ჲრკპთირვ ჲრდჲგჲპა.

Yes. It is continuous.
$f(x)=\{\begin{array}{l}x\ \text{sin}\ (x)\ \text{if}\ x\le \pi \\ x\ \text{tan}\ (x)\ \text{if}\ x>\pi \end{array},$ at $x=\pi$
$f(x)=\{\begin{array}{ll}3x+2, & x

ჲრკპთირვ ჲრდჲგჲპა.

\(k=-5$
$f(\theta )=\{\begin{array}{ll}\text{sin}\ \theta , & 0\le \theta <\frac{\pi }{2} \\ \text{cos}\ (\theta +k), & \frac{\pi }{2}\le \theta \le \pi \end{array}$
$f(x)=\{\begin{array}{ll}\frac{{x}^{2}+3x+2}{x+2}, & x\ne -2 \\ k, & x=-2\end{array}$

ჲრკპთირვ ჲრდჲგჲპა.

$k=-1$
$f(x)=\{\begin{array}{ll}{e}^{kx}, & 0\le x<4 \\ x+3, & 4\le x\le 8\end{array}$
$f(x)=\{\begin{array}{ll}\sqrt{kx}, & 0\le x\le 3 \\ x+1, & 3

ჲრკპთირვ ჲრდჲგჲპა.

\(k=\frac{16}{3}$
Let $h(x)=\{\begin{array}{ll}3{x}^{2}-4, & x\le 2 \\ 5+4x, & x>2\end{array}$ Over the interval $[0,4],$ there is no value of x such that $h(x)=10,$ although $h(0)<10$ and $h(4)>10.$ Explain why this does not contradict the IVT.
A particle moving along a line has at each time t a position function $s(t),$ which is continuous. Assume $s(2)=5$ and $s(5)=2.$ Another particle moves such that its position is given by $h(t)=s(t)-t.$ Explain why there must be a value c for $2

ჲრკპთირვ ჲრდჲგჲპა.

Since both s and \(y=t$ are continuous everywhere, then $h(t)=s(t)-t$ is continuous everywhere and, in particular, it is continuous over the closed interval $[2,5].$ Also, $h(2)=3>0$ and $h(5)=-3<0.$ Therefore, by the IVT, there is a value $x=c$ such that $h(c)=0.$

Symbols used here

\sqrt{x},\ \sqrt[n]{x}

square root, n-th root
The non-negative number whose square (n-th power) is x.

\infty

infinity
Not a number: "grows without bound" in limits and intervals.

\pi

pi
Ratio of a circle's circumference to its diameter, 3.14159…

A \cup B,\ A \cap B,\ A \setminus B

union, intersection, difference
In either; in both; in A but not B.

imaginary unit
i² = −1.

\pm

plus or minus
Both signs at once: x = 3 ± 2 means 5 and 1.

\leq,\ \geq

less/greater than or equal
Inequalities that allow equality; < and > exclude it.

\neq

not equal
The two sides are different.

\sup,\ \inf

supremum, infimum
Least upper bound, greatest lower bound.

Euler's number
2.71828…, the base whose exponential is its own derivative.

\sin,\ \cos,\ \tan

sine, cosine, tangent
Ratios of sides in a right triangle; coordinates on the unit circle.

\log_b x,\ \ln x

logarithm, natural log
The exponent b must be raised to for x; ln uses base e.

\sum_{k=1}^{n} a_k

summation
Add a_k for k = 1 up to n.

\lim_{x \to a} f(x)

limit
The value f(x) approaches as x approaches a.

f'(x),\ \frac{dy}{dx}

derivative
Instantaneous rate of change; slope of the graph.

\int f(x)\,dx,\ \int_a^b

integral
Antiderivative (indefinite) or signed area from a to b (definite).

How to: Continuity

Determine whether a function is continuous at a number.
Determine the numbers for which a function is discontinuous.
Determine whether a function is continuous.
Condition 1:
Condition 2:
Condition 3:

Questions people ask

What is a derivative in one sentence?

The slope of the graph at a point — the rate at which the output is changing there. Speed is the derivative of position.

What is an integral in one sentence?

The accumulated total of a rate — the area under the curve. Distance travelled is the integral of speed.

Why are derivatives and integrals opposites?

That is the fundamental theorem of calculus: accumulating a rate and then measuring how fast the accumulation grows gets you back the rate. Integration undoes differentiation up to a constant.

When do I use substitution and when integration by parts?

Substitution when part of the integrand is the derivative of another part (u and du both present). Parts when the integrand is a product of two unrelated kinds of function — a polynomial times an exponential, log or trig function.

ჲოთრაი ჟამ.

Parts of this page are adapted from OpenStax Calculus Volume 1 (CC BY-NC-SA 4.0), OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensed and re-explained here; errors are ours.

მეტი Calculus

Limits Derivatives Integrals Definite integrals Taylor series Series and sums Maxima and minima The chain rule Implicit differentiation Related rates and optimisation Integration techniques: substitution, parts, partial fractions Applications of integration: area, volume, arc length Infinite series and convergence tests

Polynomial functions	Ex: \(f(x)={x}^{4}-9{x}^{2}\)
Exponential functions	Ex: \(f(x)={4}^{x+2}-5\)
Sine functions	Ex: \(f(x)=\sin (2x)-4\)
Cosine functions	Ex: \(f(x)=-\cos (x+\frac{\pi }{3})\)

Logarithmic functions	Ex: \(f(x)=2\ln (x)\), \(x>0\)
Tangent functions	Ex: \(f(x)=\tan (x)+2,\) \(x\ne \frac{\pi }{2}+k\pi ,\) \(k\) is an integer
Rational functions	Ex: \(f(x)=\frac{{x}^{2}-25}{x-7},\) \(x\ne 7\)