Chapter 5.6 Continuous Functions

In this section, we focus primarily on special kinds of functions whose limits are easy to compute. We also study the properties such functions have.

Continuity

These functions are not continuous at \(x_0\):

If a function is continuous at \(x_0\), then its value at \(x_0\) is completely determined by its value at nearby points.

Finite approximation is a powerful principle for studying functions, but not all functions can be finitely approximated.

For this reason, it is important to specialize to the continuous functions.

Continous at a Point and on a Set

A real valued function \(f\) is continuous at a point \(x_0\) in \(\mathcal D(f)\) if \[\lim_{x\to x_0} f(x) = f(x_0).\]

A real valued function \(f\) is continuous on a subset \(S\) of \(\mathcal D(f)\) if it is continuous at each point of \(S\).

Understand the meaning of continuous with this example.

Example 1

Take \(f\) to be a function with the property that \[\lim_{x\to 2^-} f(x) = 5 = \lim_{x\to 2^+} f(x).\] Determine the value of \(f(2)\) so that \(f\) is continuous at \(2\).

A function is continuous at \(x_0=2\) provided that \(\lim\limits_{x\to 2}f(x)=f(2).\) This means the limit at \(x_0=2\) must equal the value of the function at \(x=2\). Since \(\lim\limits_{x\to 2^-} f(x) = 5 = \lim\limits_{x\to 2^+} f(x)\), this means the limit at \(x=2\) exists and \(\lim\limits_{x\to 2} f(x)=5.\) Therefore \(f(2)=5\) in order for \(f\) to be continuous at \(x_0=2\).

Understand the requirements of continuity with this example.

Example 2

Take \(f\) to be a function with domain equal to \([3,7]\). Given that \(f(3)\) is equal \(1\) determine the value of \(\displaystyle \lim_{x\to 3^+} f(x)\) so that \(f\) is continuous at \(3\).

A function is continuous at \(x_0=3\) provided that \(\lim\limits_{x\to 3}f(x)=f(3).\) This means the limit at \(x_0=3\) must equal the value of the function at \(x=3\). However, since the function is only defined on the interval \([3,7]\) then only limit from the right exists. Therefore, the requirement that \(\lim\limits_{x\to 3^+}f(x)=f(3)\) implies that \(\lim\limits_{x\to 3^+}f(x)=1\).

Justifying that a given function is continuous at a specified point or that it is continuous on a specified subset of its domain comes down to computing limits.

Since limits of functions are determined by sequences, the functions that are continuous at a given point \(x_0\) are precisely those that can be approximated by their values on any sequence of points that converge to \(x_0\).

This means that \(f(x_0)\) may be approximated to any degree of accuracy by studying finitely many points of \((f(x_n))\) for any sequence \((x_n)\) in \(\mathcal D(f)\smallsetminus\{x_0\}\) that converges to \(x_0\).

Example 3

Show that each function \(f\) is continuous at \(x_0\), where \(f\) and \(x_0\) are given by:

1. \(f(x) = x^2\), \(x_0 = 2\)

2. \(f(x) = \begin{cases}\dfrac{\sin(x)}{x} &\text{if } x\ne0\\1&\text{if } x= 0,\end{cases}\quad x_0=0.\)

1. A result from Chapter 5.5 for pow functions states that \(\lim\limits_{x\to x_0}x^r=x_0^r\). So in this case, \(\lim\limits_{x\to 2}x^2=2^2=4\), which is equal to \(f(2)=4.\) Thus the requirement \(\lim\limits_{x\to 2}f(x)=f(2)\) is satisfied and so \(f\) is continuous at \(x_0=2\).

2. To check continuity, calculate \(\lim\limits_{x\to 0}f(x)\). Remember, the sequential definition of limit involves sequences \((x_n)\) that converge to \(x_0=0\), but in which the terms are never equal to \(x_0=0\). So this means, the function \(f\) on \(\mathbb{R}\smallsetminus\{0\}\) is \(f(x)=\tfrac{\sin(x)}{x}\). Calculate the limit to get that \[\displaystyle\lim_{x\to 0}f(x)=\displaystyle \lim_{x\to 0} \frac{\sin(x)}{x}=1.\]Now, according to the definition of \(f\), \(f(0)=1\). Therefore, \(f\) is continuous at \(x_0=0\) because \(\lim\limits_{x\to 0}f(x)=f(0)\).

The idea of continuity may be further refined without much difficulty.

Left and Right Continuous

For any real valued function \(f\) that is defined on \(\mathbb R\), \(f\) is, respectively, left continuous or right continuous at a point \(x_0\) in its domain if \[\lim_{x\to x_0^-} f(x) = f(x_0)\quad {\rm or}\quad \lim_{x\to x_0^+} f(x) = f(x_0).\]

A function \(f\) is continuous at a point \(x_0\) in its domain if and only if it is both left and right continuous at \(x_0\).

Explanation:

\(\lim\limits_{x\to x_0}f(x)\) exists if and only if the left and right limits exist and are equal.

Practice understanding left and right continuity with the following example.

Example 4

For each real number \(x\), take \({\rm ceil}(x)=\lceil x \rceil\) to be the smallest integer that is greater than or equal to \(x\). Describe the continuity properties of \({\rm ceil}.\)

If \(-2< x\leq-1\), then \({\rm ceil}(x)=-1\).

If \(-1< x\leq 0\), then \({\rm ceil}(x)=0\).

If \(0< x\leq 1\), then \({\rm ceil}(x)=1\).

If \(1< x\leq 2\), then \({\rm ceil}(x)=2\).

Therefore, for any real number \(x_0\) such that \(Z<x_0<Z+1\) for some integer \(Z\), then \(\lim\limits_{x\to x_0}{\rm ceil}(x)={\rm ceil}(x_0).\) The ceiling function is continuous at \(x_0\) as long as \(x_0\) is not an integer.

However, the function would be left continuous at each integer \(z\) since \(\lim\limits_{x\to z^-}{\rm ceil}(x)={\rm ceil}(z).\)

Further your understanding with this example.

Example 5

For each real number \(x\), take \({\rm floor}(x)=\lfloor x\rfloor\) to be the largest integer that is smaller than or equal to \(x\). Describe the continuity properties of \({\rm floor}.\)

If \(-2\leq x<-1\), then \({\rm floor}(x)=-2\).

If \(-1\leq x<0\), then \({\rm floor}(x)=-1\).

If \(0\leq x< 1\), then \({\rm floor}(x)=0\).

If \(1\leq x<2\), then \({\rm floor}(x)=1\).

Therefore, for any real number \(x_0\) such that \(Z<x_0<Z+1\) for some integer \(Z\), then \(\lim\limits_{x\to x_0}{\rm floor}(x)={\rm floor}(x_0)\), so the ceiling function is continuous at \(x_0\) as long as \(x_0\) is not an integer.

However, the function would be right continuous at each integer \(z\) since \(\lim\limits_{x\to z^+}{\rm floor}(x)={\rm floor}(z)\)

Understand continuity of piecewise functions with the following example.

Example 6

Show that the function \(f\) is continuous at \(1\), where \(f\) is given by \[f(x) = \begin{cases}3x+2 &\text{if } x\le 1\\x+4&\text{if } x> 1.\end{cases}\]

A function \(f\) is continuous at \(x_0=1\) if \(\lim\limits_{x\to 1}f(x)=f(1)\). This means the limit at \(x_0=1\) must equal the value of the function at \(x=1\). Because \(f\) is a piecewise function that is defined differently on \((-\infty,1]\) and \((1,\infty)\), calculate \(\lim\limits_{x\to 1}f(x)\) by computing \(\lim\limits_{x\to 1^+}f(x)\) and \(\lim\limits_{x\to 1^-}f(x).\)

When \(x<1\), \(f(x)=3x+2\). Therefore, \[ \begin{align*} \lim_{x\to 1^-}f(x)&=\lim_{x\to 1^-}(3x+2)\\ &=\lim_{x\to 1^-}3\cdot \lim_{x\to 1^-}x+\lim_{x\to 1^-}2\\ &=3\cdot1+2\\ &=5. \end{align*} \]

When \(x>1\), \(f(x)=x+4\). Therefore, \[ \begin{align*} \lim_{x\to 1^+}f(x)&=\lim_{x\to 1^+}(x+4)\\ &=\lim_{x\to 1^+}x+\lim_{x\to 1^+}4\\ &=1+4\\ &=5. \end{align*} \]

Since \(\lim\limits_{x\to 1^+}f(x)=\lim\limits_{x\to 1^-}f(x)\), the limit exists at \(x_0=1\) and \(\lim\limits_{x\to 1}f(x)=5\). Moreover, \(f(1)=3\cdot 1+2=5\).

Because \(\lim\limits_{x\to 1}f(x)=f(1)\), the function is continuous at \(x_0=1.\)

Further your understanding of continuity of piecewise functions with this example.

Example 7

Show that the function \(f\) is continuous at \(0\), where \(f\) is given by \[f(x) = \begin{cases}x^2-5x &\text{if } x\le 0\\x\sin\left(\tfrac{1}{x}\right)&\text{if } x> 0.\end{cases}\]

A function is continuous at \(x_0=0\) if \(\lim\limits_{x\to 0}f(x)=f(0)\). This means \(x_0=0\) must equal the value of the function at \(x=0\). Because \(f\) is a piecewise function that is defined depending on if \(x\leq 0\) or if \(x>0\), calculate \(\lim\limits_{x\to 0}f(x)\) by computing \(\lim\limits_{x\to 0^+}f(x)\) and \(\lim\limits_{x\to 0^-}f(x).\)

When \(x<0\), \(f(x)=x^2-5x\). Therefore, \[ \begin{align*} \lim_{x\to 0^-}f(x)&=\lim_{x\to 0^-}(x^2-5x)\\ &=0. \end{align*} \]

When \(x>0\), \(f(x)=x\sin\left(\frac{1}{x}\right)\). Therefore, \[ \begin{align*} \lim_{x\to 0^+}f(x)&=\lim_{x\to 0^+}x\sin\left(\frac{1}{x}\right)\\ &=0&&\text{by the squeeze theorem.} \end{align*} \]

Since \(\lim\limits_{x\to 0^+}f(x)=\lim\limits_{x\to 0^-}f(x)\), the limit exists at \(x_0=0\) and \(\lim\limits_{x\to 0}f(x)=0\). Moreover, \(f(0)=0\).

Because \(\lim\limits_{x\to 0}f(x)=f(0)\), the function is continuous at \(x_0=0.\)

A function may not necessarily be continuous at a point because it is undefined. Depending on the behave of the function near the place it is undefined, there may be a way to construct a continuous function out of it.

Continuous Extension of a Function

For any real valued function \(f\) that is defined on a subset of \(\mathbb R\), if \(x_0\) is not in the domain of \(f\), but there is a continuous function \(g\) that is defined at \(x_0\) and for every \(x\) in \(\mathcal D(f)\), \[f(x) = g(x),\] then \(g\) is a continuous extension of \(f\) to \(x_0\).

Here is how a function can have a continuous extension.

Admits a Continuous Extension

For any real valued function \(f\) that is defined on a subset of \(\mathbb R\), if \(x_0\) is not in the domain of the function \(f\), but there is some real number \(L\) so that \[\lim_{x\to x_0}f(x) = L,\] then \(f\) admits a continuous extension to the set \(\mathcal D(f)\cup\{x_0\}\)

To say that \(f\) admits a continuous extension to the set \(\mathcal D(f)\cup\{x_0\}\) means that there is a continuous extension \(g\) of \(f\) to \(x_0\), and that \(g\) is defined by \[g(x) = \begin{cases} f(x)&\text{if } x\in \mathcal D(f)\\L&\text{if } x = x_0.\end{cases}\]

One type of discontinuity a function can have is called a removable singularity.

Removable Singularity

If \(x_0\) is a limit point of \(\mathcal D(f)\), then the continuous extension of \(f\) to \(x_0\), if it exists, is unique and \(x_0\) is a removable singularity of \(f\).

Practice finding a continuous extension with this example.

Example 8

Take \(f\) to be the function that is given by \[f(x) = \tfrac{2x^2 -x -1}{x-1}.\] Determine the maximal domain of \(f\) and find a continuous extension of \(f\) to all of \(\mathbb R\).

The maximal domain of \(f\) is \((-\infty,1)\cup(1,\infty)\) because the function is undefined only at \(x=1\). However, the function does have a limit at \(x=1\):

\[ \begin{align*} \lim_{x\to 1}f(x)&=\lim_{x\to 1}\tfrac{2x^2 -x -1}{x-1}\\ &=\lim_{x\to 1}\tfrac{(2x+1)(x-1)}{x-1}\\ &=\lim_{x\to 1}(2x+1)\\ &=\lim_{x\to 1}2\cdot \lim_{x\to 1}x+\lim_{x\to 1}1\\ &=2\cdot 1+1\\ &=3. \end{align*} \]

If the function is defined to equal \(3\), when \(x=1\), then it would be continuous. Define the function \(g\) as:

\[ g(x)=\begin{cases} \tfrac{2x^2 -x -1}{x-1}&\text{ if }x\not=1\\ 3 &\text{ if } x=1. \end{cases} \]

Then \(g\) is a continuous extension of \(f\) to all of \(\mathbb R\) since \(g(x)=f(x)\) when \(x\) in \((-\infty,1)\cup(1,\infty)\).

Not every function has a continuous extension. Use this example to understand why.

Example 9

Take \(f\) to be the function that is given by \[f(x) = \tfrac{|x-2|}{x-2}.\] Determine the domain of \(f\) and show that there is no continuous extension of \(f\) to all of \(\mathbb R\).

The domain of \(f\) is \((-\infty,2)\cup(2,\infty)\). There is no continuous extension of \(f\) because the limit does not exist at \(2\) since \(\lim\limits_{x\to 2^-}f(x)=-1\) and \(\lim\limits_{x\to 2^+}f(x)=1\) are not equal.

What about functions that are not defined on all real numbers? Understand the subtlety with this example.

Example 10

Recall that \({\rm pow}_{\frac{1}{2}}\) is the function that is given by \[{\rm pow}_{\frac{1}{2}}(x) = \sqrt{x}.\]

1. Determine the maximal domain of \({\rm pow}_{\frac{1}{2}}\) and show that \({\rm pow}_{\frac{1}{2}}\) is continuous on this domain.

2. Find a continuous extension of \({\rm pow}_{\frac{1}{2}}\) to all of \(\mathbb R\).

3. Is the continuous extension of \({\rm pow}_{\frac{1}{2}}\) to all of \(\mathbb R\) unique?

4. A common error that students make is to assert that \({\rm pow}_{\frac{1}{2}}\) is not continuous at \(0\). Why do they make this mistake?

1. The maximal domain is \([0,\infty)\). The function is continuous on this domain because \(\displaystyle\lim_{x\to x_0}\sqrt{x}=\sqrt{x_0}\) for \(x_0\) in \([0,\infty)\).

2. A continuous extension is \[g(x)=\begin{cases}\sqrt{x}&\text{ if }x\geq 0\\0&\text{ if }x<0.\end{cases}\] since \(g\) is continuous on \(\mathbb{R}\) and \(\sqrt{x}=g(x)\) when \([0,\infty)\).

3. No, here is another one \[h(x)=\begin{cases}\sqrt{x}&\text{ if }x\geq 0\\x^2&\text{ if }x<0.\end{cases}\] since \(g\) is continuous on \(\mathbb{R}\) and \(\sqrt{x}=h(x)\) when \([0,\infty)\).

4. Because \(\sqrt{x}\) is undefined when \(x<0\). However, when computing limits, the sequences are defined only on the domain of \(f\).

The following definition clarifies what it means for a function to not be continuous at a point.

Not Continuous at a Point

A function \(f\) is not continuous at a point \(x_0\) in \(\mathcal D(f)\) if \(x_0\) is a limit point of \(\mathcal D(f)\) and either

• \(\displaystyle \lim_{x\to x_0}f(x)\) does not exist;

or

• \(\displaystyle \lim_{x\to x_0}f(x)\) exists but is not equal to \(f(x_0)\).

A function \(f\) may also fail to be continuous at a point \(x_0\) because \(x_0\) is not in the domain of \(f\).

Note that if \(x_0\) is an isolated point of \(\mathcal D(f)\), then \(f\) is continuous at \(x_0\).

Do you see why it is “vacuously true” that functions are continuous on isolated points in their domain?

Check your understanding with the next example.

Example 11

Carefully show that the function \(f\) is not continuous at \(x_0\), where \(f\) and \(x_0\) are given by:

1. \(f(x) = \begin{cases} 3x+1 & \text{if } x < 2\\ x-5 & \text{if } x\ge 2,\end{cases}\quad x_0=2\)

2. \(f(x) = \begin{cases} x^2 & \text{if } x \ne 3\\ 5 & \text{if } x=3,\end{cases}\quad x_0=3\)

3. \(f(x) = \begin{cases} \frac{1}{x^2} & \text{if } x \ne 0\\ 0 & \text{if } x=0,\end{cases}\quad x_0 = 0\)

4. \(f(x) = \begin{cases} 0 & \text{if } x\le 0\\ \sin\!\big(\frac{1}{x}\big) & \text{if } x> 0,\end{cases}\quad x_0 = 0\)

1. The function \(f\) is a piecewise function defined depending on if \(x<2\) or \(x\geq 2\). Calculate the limit at \(x_0=2\) by calculating the limit from the left and from the right. The limits are \[\begin{align*}\displaystyle \lim_{x\to 2^-}f(x)=\displaystyle \lim_{x\to 2^-}(3x+1)=7\end{align*}\] and \[\begin{align*}\displaystyle \lim_{x\to 2^+}f(x)=\displaystyle \lim_{x\to 2^+}(x-5)=-3\end{align*}.\] Since the one-sided limits are not equal, \(f\) has no limit at \(x=2\). Therefore, \(f\) is not continuous at \(x=2\).

2. The function \(f\) is a piecewise function defined differently depending on if \(x=3\) or \(x\not= 3\). Therefore, calculate the limit at \(3\) directly. The limit is \[\begin{align*}\displaystyle \lim_{x\to 3}f(x)=\displaystyle \lim_{x\to 3}x^2=9\end{align*}.\] However, the value of the function at \(x=3\) is \(f(3)=5\) which does not equal the limit at \(x=3\). Therefore \(f\) is not continuous at \(x=3\).

3. The function \(f\) is a piecewise function defined differently depending on if \(x=0\) or \(x\not= 0\). Calculate the limit at \(x_0=0\) directly. The limit is \[\begin{align*}\displaystyle \lim_{x\to 0}f(x)=\displaystyle \lim_{x\to 0}\frac{1}{x^2}=\infty\end{align*}.\] Because the limit diverges to infinity, the function \(f\) is not continuous at \(x=0\).

4. The function \(f\) is a piecewise function defined differently depending on if \(x<0\) or \(x\geq 0\). Calculate the limit at \(x_0=0\) by calculating the limit from the left and from the right. The limit on the left is \[\begin{align*}\displaystyle \lim_{x\to 0^-}f(x)=\displaystyle \lim_{x\to 0^-}0=0\end{align*}.\] The limit on the right \[\begin{align*}\displaystyle \lim_{x\to 0^+}f(x)=\displaystyle \lim_{x\to 0^+}\sin\left(\tfrac{1}{x}\right)\end{align*}\] does not exist. Because \(f\) has no limit at \(x=0\), the function \(f\) is not continuous at \(x=0\).

There are actually many functions that are continuous that we have already studied before.

Theorem on Continuous Function

The limit laws imply that rational functions are continuous wherever they are defined.

The functions \(\sin\) and \(\cos\) are continuous on \(\mathbb R\). The continuity of the \(\sin\) and \(\cos\) functions together with the limit laws imply that the functions \(\tan\), \(\sec\), \(\csc\), and \(\cot\) will also be continuous wherever they are defined.

The absolute value function is continuous at every \(x\) in \(\mathbb R\).

The exponential function \(\text{exp}_a(x)=a^x\) is continuous at every \(x\) in \(\mathbb R\).

The logarithm function \(\text{log}_a(x)\) is continuous at every \(x\) in \((0,\infty)\).

Finally, for every natural number \(n\), the function \({\rm pow}_{\frac{1}{n}}\) is continuous at every \(x\) in its domain.

Use these properties to simplify computing limits!

The final thing we will discuss is continuity for a path. A path \(c\) in \(\mathbb{R}^2\) is continuous at a point \(t_0\) if \(t_0\) is in \(\mathcal{D}(c)\) and \[\lim_{t\to t_0}c(t)=c(t_0).\]

This is equivalent to requiring that the component functions of \(c\) are continuous at \(t_0.\)

Example 12

Take \(c\) to be a path that is given by

\[ c(t)=\begin{cases}(2t-1,3t+5)&\text{if }t\leq 1\\(5t-4,t+3)&\text{if }t>1.\end{cases} \] Determine whether \(c\) is continuous at \(1.\)

First determine the limit at \(1\) by calculating the left and right hand limit:

\[ \lim_{t\to 1^-}c(t)=\left(\lim_{t\to 1^-}(2t-1),\lim_{t\to 1^-}(3t+5)\right)=(1,8) \] and

\[ \lim_{t\to 1^+}c(t)=\left(\lim_{t\to 1^+}(5t-4),\lim_{t\to 1^+}(t+3)\right)=(1,4). \] The limits are not equal, so \(c\) is not continuous at \(1.\)

Properties of Continuous Functions

Continuous functions are very special kinds of functions. In this section, we learn all about their special properties. First, we discuss maximum and minimum values.

Attain a Maximum or Minimum Value

For any real valued function \(f\) that is defined on a subset \(S\) of \(\mathbb R\), \(f\) attains its maximum value on \(S\) if there is an \(x_0\) in \(S\) so that for any \(x\) in \(S\), \[f(x) \leq f(x_0).\]

Similarly, \(f\) attains its minimum value on \(S\) if there is an \(x_0\) in \(S\) so that for any \(x\) in \(S\), \[f(x) \geq f(x_0).\]

This theorem tells us that a continuous function will attain its maximum and minimum on a closed and bounded interval.

Extreme Value Theorem

The Extreme Value Theorem states the following:

For any function \(f\) that is continuous on a closed and bounded interval \([a,b]\), \(f\) attains its maximum and minimum values on \([a,b]\).

Understand under which circumstances the extreme value theorem applies with the following example.

Example 13

Take \(f\) to be the function that is given by \[f(x) = \tfrac{1}{x}.\] For each subset \(S\) of the domain of \(f\), say whether or not \(f\) attains its minimum value on \(S\), its maximum value on \(S\), or if it attains neither its minimum nor maximum value on \(S\):

1. \(S = (-\infty, 0)\);

2. \(S = (0, \infty)\);

3. \(S = (-\infty, -2]\);

4. \(S = [1,\infty)\).

1. The function \(f\) attains neither its minimum or maximum value on \(S\).

2. The function \(f\) attains neither its minimum or maximum value on \(S\).

3. The function \(f\) attains its minimum at \(x=-2\), but it does not attain its maximum.

4. The function \(f\) attains its maximum at \(x=1\), but it does not attain its minimum.

Continuous functions are special because they have the following property.

Intermediate Value Property

A real valued function \(f\) that is defined on the set \(S\) has the intermediate value property if, given points \(a\) and \(b\) in \(S\) with \[f(a) < f(b),\] for any \(z\) in \((f(a), f(b))\), there is a \(c\) in \(S\) that is between \(a\) and \(b\) so that \[f(c) = z.\]

This is next theorem tells us that continuous functions have the intermediate value property.

Intermediate Value Theorem

The Intermediate Value Theorem is the following statement:

Every continuous real valued function \(f\) that is defined on an interval \(I\) has the intermediate value property.

Understand the meaning with this next example.

Example 14

Depict graphically the meaning of the intermediate value theorem

Given any interval \(I=[a,b],\) the continuous function \(f\) will take on all values in the interval \([f(a),f(b)]\) or \([f(b),f(a)]\) depending on if \(f(a)<f(b)\) or \(f(b)<f(a).\) So this means if \(z\) is in \([f(a),f(b)],\) then there is a \(c\) in \(I\) so that \(f(c)=z.\)

The following theorem restates the above property in terms of the range of a continuous function.

Continuous Function Take Closed and Bounded Intervals to Closed and Bounded Intervals

For any function \(f\) that is continuous on a closed and bounded interval \([a,b]\), the extreme value theorem and the intermediate value theorem together imply that there are real numbers \(m\) and \(M\) so that \[f([a,b]) = [m, M].\]

We summarize this fact by saying that continuous functions take closed and bounded intervals to closed and bounded intervals.

Note that the intermediate value theorem implies that continuous functions take intervals to intervals, but it is certainly possible that a continuous function takes an open interval to open interval, a half-open interval, or a closed interval.

So what about continuous paths? Do they have an intermediate value theorem? To understand what that means in \(\mathbb{R}^2,\) we need to further study lines in the plane.

For any line \(L\) in \(\mathbb{R}^2\) and any points \(p_1\) and \(p_2\) in \(\mathbb{R}^2\) that are not in \(L\), there are points \(q_1\) and \(q_2\) in \(L\) that are, respectively, the closest point in \(L\) to \(p_1\) and the closest point in \(L\) to \(p_2.\)

Since \(p_1-q_1\) and \(p_2-q_2\) are non-zero and perpendicular to any vector that moves points along \(L\), there is a real number \(c\) so that \[c\cdot (p_1-q_1)=p_2-q_2.\]

The points \(p_1\) and \(p_2\) lie on opposite sides of \(L\) if \(c\) is negative, and they lie on the same side of \(L\) if \(c\) is positive. So what this means is that lines divide the plane into two parts.

Understand this idea with the following example.

Example 15

Take \(L\) to be the line with slope \(\tfrac{1}{2}\) that intersects \((1,3).\)

1. Show that the points \((-1,3)\) and \((4,2)\) lie on opposite sides of \(L.\)
2. Show that the points \((-1,3)\) and \((2,6)\) lie on the same side of \(L\).

The line \(L\) is given by the equation \(y=\tfrac{1}{2}(x-1)+3.\) To determine which side the points line on, identify the points on \(L\) that are closest to the given points. The point \((3,4)\) is on \(L\) and the closest to both \((2,6)\) and \((4,2).\) The point \(\left(-\tfrac{3}{5},\tfrac{11}{5}\right)\) is the point on \(L\) that is closest to \((-1,3).\)

So the vectors that take you from the closest points on \(L\) to the points off the line are

\[ V_1=(-1,3)-\left(-\tfrac{3}{5},\tfrac{11}{5}\right)=\left\langle -\tfrac{2}{5},\tfrac{4}{5}\right\rangle, \]

\[ V_2=(2,6)-(3,4)=(-1,2), \]

and

\[ V_3=(4,2)-(3,4)=(1,-2). \]

Their respective unit vectors are

\[ \hat{V_1}=\tfrac{1}{\sqrt{5}}\langle -1,2\rangle, \quad \hat{V_2}=\tfrac{1}{\sqrt{5}}\langle -1,2\rangle,\quad\text{and}\quad \hat{V_3}=-\tfrac{1}{\sqrt{5}}\langle 1,2\rangle. \]

Since \(\hat{V_1}=\hat{V_2}\), the points \((-1,3)\) and \((2,6)\) line on the same side of \(L\). Since \(\hat{V_1}=-\hat{V_3},\) the points \((-1,3)\) and \((4,2)\) lie on opposite sides of \(L\).

Now we state what we mean by the intermediate value theorem for continuous paths.

Intermediate Value Theorem for Continuous Paths

For any line \(L\) in \(\mathbb{R}^2\) and any path \(c\) with domain \([a,b]\), if

• \(c\) is continuous on \([a,b]\),

• and \(c(a)\) and \(c(b)\) lie on opposite sides of \(L\),

then there is a \(t_0\) in \((a,b)\) so that \(c(t_0)\) is in \(L.\)

What this says is that continuous paths in the plane cannot jump over lines. Understand this idea with the next example.

Example 16

Take \(L\) to be the line that is given by

\[ L=\big\{(3t+2,t+1)\colon\, t\in\mathbb{R}\big\}. \]

Take \(c\) to be the path with domain \(\mathbb{R}\) that is given by

\[ c(t)=(t^3-t+1,2t^2). \]

Use the intermediate value theorem for paths to show that \(c\) must intersect \(L.\)

The path \(c\) evaluated at \(0\) and \(1\) is at \(c(0)=(1,0)\) and \(c(1)=(1,2).\) The point \(\left(1,\tfrac{2}{3}\right)\) is in \(L.\) The vector \(V=\langle 3,1\rangle\) moves points along \(L\) and the vector \(\hat{V_\perp}=\tfrac{1}{\sqrt{10}}\langle -1,3\rangle\) moves points in a direction that is perpendicular to \(L.\) Notice that

\[ \left(c(0)-\left(1,\tfrac{2}{3}\right)\right)\cdot \hat{V_\perp}=\left\langle 0,-\tfrac{2}{3}\right\rangle\cdot\tfrac{1}{\sqrt{10}}\langle -1,3\rangle =\tfrac{1}{\sqrt{10}}(-2)<0 \]

and

\[ \left(c(1)-\left(1,\tfrac{2}{3}\right)\right)\cdot \hat{V_\perp}=\left\langle 0,\tfrac{4}{3}\right\rangle\cdot\tfrac{1}{\sqrt{10}}\langle -1,3\rangle =\tfrac{1}{\sqrt{10}}(4)>0. \]

Take \(p_0\) to be the point in \(L\) that is closest to \(c(0)\) and \(p_1\) to be the point in \(L\) that is closest to \(c(1).\)

Since \(c(0)-p_0=\tfrac{1}{\sqrt{10}}(-2)\cdot \hat{V_\perp}\) and \(c(1)-p_1=\tfrac{1}{\sqrt{10}}(4)\cdot \hat{V_\perp}\) point in opposite directions, \(c(0)\) and \(c(1)\) lie on opposite sides of \(L.\) Since \(c\) is continuous, there is a \(t_0\) in \((0,1)\) so that \(c(t_0)\) is in \(L\).

Here are the equivalent limit laws for continuous functions.

Continuity Laws: Sum, Products, Polynomials and Rational Functions

The limit laws immediately imply that if \(f\) and \(g\) are both defined and continuous on a set \(S\), then \(fg\) and \(f+g\) are both continuous on \(S\).

If in addition, \(g\) is nonzero on \(S\), then the quotient \(\frac{f}{g}\) is also continuous on \(S\).

Since the function \({\rm pow}_1\) is continuous on \(\mathbb R\), the limit laws therefore imply that polynomials are continuous.

Since polynomial functions are continuous, rational functions are also continuous wherever they are defined.

Practice using the continuity laws with the following examples.

Example 17

For each function \(f\), decompose \(f\) into sums, products, and quotients of continuous functions to find a set \(S\) on which \(f\) is continuous:

1. \(f(x) = \frac{(x-1)(x+1)^3}{x^4 +3}\)

2. \(f(x) = \frac{(\cos(x)-5)(x^2+1)}{x\cos(x)} + \tan(x)\)

1. Take \(a(x)=(x-1)\), \(b(x)=(x+1)\), and \(c(x)=x^4+3\). Then \(f\) decomposes like this: \(f=\frac{a\cdot b\cdot b\cdot b}{c}\). The function will be continuous on \(S\) so long as the numerator is continuous on \(S\), the denominator is continuous on \(S\), and the denominator is not equal to zero.
   • The numerator, \(a\cdot b\cdot b\cdot b\) is the product of functions, so it will be continuous so long as all functions are continuous on the same set. Since \(a=(x-1)\) and \(b=(x+1)\) are polynomials, they are continuous on \(\mathbb{R}\). Therefore, the numerator is continuous on \(\mathbb{R}\).
   • The denominator, \(c\) is a polynomial, so it is continuous on \(\mathbb{R}\).
   • The denominator, \(c(x)=x^4+3\), is zero when \(x^4+3=0\). There is no solution, so we do not need to exclude any numbers in our answer. So the common set \(S\) is \(\mathbb{R}\).
2. Take \(a(x)=\cos(x)\), \(b(x)=x^2+1\), \(c(x)=x\), \(d(x)=5\), and \(e(x)=\tan(x)\). Then \(f\) decomposes like this: \(f=\frac{(a-d)\cdot b}{c\cdot a}+e\). The function will be continuous on \(S\) so long as the numerator is continuous on \(S\), the denominator is continuous on \(S\), and the denominator is not equal to zero, and \(e\) is continuous on \(S\)
   • The numerator: \((a-d)\cdot b\) is the product of functions, so it will be continuous so long as all functions are continuous on the same set. The function \((a-d)(x)=\cos(x)-5\) is a cosine function shifted down by 5. The function cosine is continuous on all real numbers, so \(a-d\) will still be continuous on all real numbers. The function \(b(x)=x^2+1\) is a polynomial so it is also continuous on the same set. Therefore, the numerator is continuous on \(\mathbb{R}\).
   • The denominator: \(c\) is a polynomial, so it is continuous on \(\mathbb{R}\). \(a\) is is the cosine function so it is likewise continuous on \(\mathbb{R}\). Therefore, the denominator is continuous on \(\mathbb{R}\).
   • The denominator, \(x\cos(x)\), is zero when \(x\cos(x)=0\). One solution is \(x=0\) and the other is when \(\cos(x)=0\). So we exclude \(x=0\) and \(x=\tfrac{\pi}{2}k\), where \(k\) is an odd integer.
   • The term \(e(x)=\tan(x)\) is continuous as long as \(x\) is not equal to \(x=\tfrac{\pi}{2}k\), where \(k\) is an odd integer. So the common set \(S\) is \(\mathbb{R}\smallsetminus \left\{0,\tfrac{\pi}{2}k\right\}\) or \(\left\{x\in\mathbb{R}\colon x\not =0 \text{ or } x\not= \tfrac{\pi}{2}k , k\text{ is an odd integer}\right\}.\)

The situation for composite functions is a little more subtle. Here is the theorem for composite functions.

Continuity for Composite Functions

Take \(g\) to be a function on an interval \(I\) in \(\mathbb R\) that is continuous at a point \(x_0\) in \(I\).

Take \(f\) to be a function on \(g(I)\).

If \(f\) is continuous at \(g(x_0)\), then \(f\circ g\) is continuous at \(x_0\).

Here is a proof of the above fact.

Example 18

Give a proof of the above fact.

Take a sequence \((x_n)\) in \(\mathcal{D}(g)\smallsetminus \{x_0\}=I \smallsetminus\{x_0\}\) such that \(\lim\limits_{n\to\infty}x_n=x_0\). Show \(\lim\limits_{n\to\infty}(f\circ g)(x_n)=\lim\limits_{n\to\infty}f(g(x_n))\) converges to \(f(g(x_0))\). Because \(g\) is continuous at \((x_0)\) and \(f\) is continuous at \(g(x_0)\),

• \(\lim\limits_{n\to\infty}g(x_n)=g(x_0)\)

• \(\lim\limits_{n\to\infty}f(g(x_n))=f(g(x_0)).\)

Therefore \(\lim\limits_{n\to\infty}(f\circ g)(x_n)=f(g(x_0)).\)

If we are just interested in calculating limits of composites, then we just need \(g\) to have a limit \(L\) and the function \(f\) to be continuous at \(L\).

Example 19

Write each function \(h\) that is given below as a compound function to determine \(\lim\limits_{x\to x_0}h(x)\), where \(h\) and \(x_0\) are given by:

1. \(h(x) = \sin\left(\frac{x^2 -x - 2}{x-2}\right)\), \(x_0 = 2\)
2. \(h(x) = \sqrt{\frac{1-\cos(x-1)}{x-1}}\), \(x_0 = 1\)
3. \(h(x) = \exp_3\left(\frac{x^2 -x - 2}{x+1}\right)\), \(x_0 = -1\)
4. \(h(x) = \log_5\left(\frac{\sin(2x)}{3x}\right)\), \(x_0 = 0\)

1. The function can be rewritten as \(h=\sin\circ g\) where \(g(x)=\tfrac{x^2 -x - 2}{x-2}\). Check that \(\lim\limits_{x\to 2}g(x)\) exists and \(\sin\) is continuous at that value to determine the limit of \(h\):\[\begin{align*}\lim_{x\to 2}g(x)&=\lim_{x\to 2}\frac{x^2-x-2}{x-2}\\ &=\lim_{x\to 2}\frac{(x-2)(x+1)}{x-2}\\ &=\lim_{x\to 2}(x+1)\\ &=3.\end{align*}\] Since \(\sin\) is continuous on \(\mathbb{R}\) it is continuous at \(3\). Therefore by above theorem; \[\displaystyle \lim_{x\to 2}\sin\left(\frac{x^2 -x - 2}{x-2}\right)=\sin\left(\displaystyle \lim_{x\to 2}\frac{x^2 -x - 2}{x-2}\right)=\sin(3).\]

2. The function can be rewritten as \(h={\mathrm{pow}}_\frac{1}{2}\circ g\) where \(g(x)=\tfrac{1-\cos(x-1)}{x-1}\). Check that \(\lim\limits_{x\to 1}g(x)\) exists and that \({\mathrm{pow}}_\frac{1}{2}\) is continuous at that value to determine the limit of \(h\): \[\begin{align*}\lim_{x\to 1}g(x)&=\lim_{x\to 1}\frac{1-\cos(x-1)}{x-1}\\ &=0 \end{align*}\] since \(\lim\limits_{\star \to 0}\tfrac{1-\cos(\star)}{\star}=1\) from Chapter 5.3. Since \(\mathrm{pow}_\frac{1}{2}\) is continuous on \([0,\infty),\) it is continuous at \(0\). Therefore by above theorem; \[\displaystyle \lim_{x\to 1}\sqrt{\frac{1-\cos(x-1)}{x-1}}=\sqrt{\lim_{x\to 1}\frac{1-\cos(x-1)}{x-1}}=\sqrt{0}=0.\]

3. The function can be rewritten as \(h=\exp_3\circ g\) where \(\exp_3(x)=3^x\) and \(g(x)=\tfrac{x^2 -x - 2}{x+1}\). We need to check that \(\lim\limits_{x\to -1}g(x)\) exists and that \(f\) is continuous at that value to conclude what the limit of \(h\) will be: \[\begin{align*}\lim_{x\to -1}g(x)&=\lim_{x\to -1}\frac{x^2 -x - 2}{x+1}\\ &=\lim_{x\to -1}\frac{(x-2)(x+1)}{x+1}\\ &=\lim_{x\to -1}(x-2)\\ &=-3. \end{align*}\] Since \(\exp_3\) is continuous on \(\mathbb{R},\) it is continuous at \(-3\). Therefore by the above theorem; \[\displaystyle \lim_{x\to -1}\exp_3\left(\frac{x^2 -x - 2}{x+1}\right)=\exp_3\left(\lim_{x\to -1}\frac{x^2 -x - 2}{x+1}\right)=\exp_3(-3)=3^{-3}=\frac{1}{27}.\]

4. The function can be rewritten as \(h=\log_5\circ g\) where \(g(x)=\tfrac{\sin(2x)}{3x}\). Check that \(\lim\limits_{x\to 0}g(x)\) exists and that \(f\) is continuous at that value to determine the limit of \(h\): \[\begin{align*}\lim_{x\to 0}g(x)&=\lim_{x\to 0}\frac{\sin(2x)}{3x}\\ &=\frac{2}{3}. \end{align*}\] Since \(\log_5\) is continuous on \((0,\infty),\) it is continuous at \(\tfrac{2}{3}\). Therefore by the above theorem; \[\displaystyle \lim_{x\to 0}\log_5\left(\frac{\sin(2x)}{3x}\right)=\log_5\left(\lim_{x\to 0}\frac{\sin(2x)}{3x}\right)=\log_5\left(\tfrac{2}{3}\right).\]

Let’s use what we know to construct functions with certain properties.

Example 20

Construct a function \(f\) that is continuous everywhere except at \(3,\) that is left continuous at \(3,\) that is strictly decreasing to the right of \(3,\) that is asymptotically equal to \(-5x^\frac{1}{3}\) to the right, strictly decreasing to the left of \(3\) and asymptotically equal to \(4x^2\) to the left, and has the property that \[\lim_{x\to 3^+}f(x)=4\quad\text{and}\quad \lim_{x\to 3^-}f(x)=-3.\]

We can construct this function by creating a piecewise function.

For \(x\leq 3,\) we want the function to be left continuous at \(3\), strictly decreasing, asymptotically equal to \(4x^2\), and the limit from the left at \(3\) should be \(-3.\) So \(g(x)=4(x-3)^2-3\) works for \(x\leq 3.\)

For \(x>3\), we want the function to be strictly decreasing, asympotically equal to \(-5x^\frac{1}{3}\), and the limit from the right at \(3\) should be \(4.\) So \(h(x)=-5(x-3)^\frac{1}{3}+4\) works for \(x>3\).

Thus the function is

\[ f(x)= \begin{cases} 4(x-3)^2-3&\text{if }x\leq 3\\ -5(x-3)^\frac{1}{3}+4&\text{if }x>3. \end{cases} \]

Example 21

Construct a continuous path \(c\) with domain \([0,\infty)\) that describes the position of a particle that moves to the right on the line segment from \((2,7)\) to \((8,-3)\) is at \((2,7)\) at time \(0\), is never at the same point at different time points, that never reaches \((8,-3),\) but that has the property that \[ \lim_{t\to\infty}\|(8,-3)-c(t)\|=0. \]

Take \(V=(8,-3)-(2,7)=\langle 6,-10\rangle.\) Our path will be of the form \(c(t)=g(t)\langle 6,-10\rangle +(2,7)\) where \(g\) is a function that satifies the following:

• \(g(0)=0\) so that \(c(0)=(2,7),\)
• \(g\) is strictly increasing on \([0,\infty)\) so that it is never at the same point at different time points,
• \(\lim\limits_{t\to\infty}g(t)=1\), meaning \(g\) is asymptotically equal to \(1\) so \(c\) approaches \((8,-3)\) as \(t\) increases. Here is one example for \(g\): \[g(t)=\frac{t}{t+1}.\] Hence our path is \[c(t)=\tfrac{t}{t+1}\langle 6,-10\rangle +(2,7).\]

Example 22

Construct a continuous path \(c\) with domain \([0,\infty)\) that describes the position of a particle that rotates clockwise around a circle of radius \(3\) that is centered at \((1,5)\), that is at \((4,5)\) at time \(0\), that always approaches \((4,5)\) after leaving \((4,5)\) and gets as close as we like to \((4,5)\), but never returns to \((4,5).\)

This path is essentially rotation on a circle of radius \(3\) with center \((1,5)\) with certain requirements. So, it will look something like this:

\[ c(t)=\left(\cos(g(t)),\sin(g(t))\right)\star \langle 3,0\rangle +(1,5), \] where \(g\) is a function that satisfies the following requirements:

• \(g(0)\) so that \(c(0)=(4,5),\)
• \(g\) is strictly increasing on \([0,\infty)\) so it never returns to \((4,5),\)
• \(\lim\limits_{t\to\infty}g(t)=2\pi\) so that it approaches one full rotation which means it approaches \(c\) approaches the point \((4,5).\) Here is one example for \(g\): \[g(t)=\frac{2\pi t}{t+1}.\] Hence our path is \[ c(t)=\left(\cos\left(\tfrac{2\pi t}{t+1}\right),\sin\left(\tfrac{2\pi t}{t+1}\right)\right)\star \langle 3,0\rangle +(1,5). \]

Approximating Continuous Functions

The Intermediate Value Theorem is an existence theorem, it guarantees that certain equations that involve continuous functions have solutions.

For example, if \(f\) and \(g\) are continuous functions that are defined on the same interval \([a,b]\) and if \(f(a)-g(a)\) and \(f(b)-g(b)\) have different signs, then there is at least one solution to the equation \[f(x) = g(x)\] on \([a,b]\).

Knowing that there is a solution gives a way of finding a solution to the equation.

Example 23

Show that the equation \[x = \cos(x)\] has at least one solution.

Set \(f(x)=x\) and \(g(x)=\cos(x)\). Both functions are continuous on all of \(\mathbb{R}\) because \(f\) is a polynomial and \(g\) is the cosine function. Focus on \(f(x)-g(x)=x-\cos(x)\).

Notice that when \(x=0\), \(f(0)-g(0)=-1\) and when \(x=\frac{\pi}{2}\), \(f(\frac{\pi}{2})-g(\frac{\pi}{2})=\frac{\pi}{2}\). On the interval \([0,\frac{\pi}{2}]\), \(f(0)-g(0)\) and \(f(\frac{\pi}{2})-g(\frac{\pi}{2})\) have different signs.

Because \(h(x)=f(x)-g(x)=x-\cos(x)\) is continuous on \([0,\frac{\pi}{2}]\) and \(h(0)<0\) and \(h(\frac{\pi}{2})>0\), the Intermediate Value Theorem states that there exists a value \(z\) in \([0,\frac{\pi}{2}]\), for which \(h(z)=0.\)

In other words, \(z-\cos(z)=0\) or \(z=\cos(z)\). So there is at least one solution to the equation \[x = \cos(x).\]

In this example, pay attention to how the intermediate value theorem is used.

Example 24

Take \(f\) to be the polynomial that is given by \[f(x) = x^7 - 4x^3 +x^2 - 3.\] Show that \(f\) has at least one real root.

Use the intermediate value theorem by finding an interval \([a,b]\), so that \(f(a)\) and \(f(b)\) have opposite sign.

Focus on \([-1,0]\). Since \(f\) is a polynomial, it is continuous everywhere, including the interval \([-1,0]\). Notice that \(f(-1)=1\) while \(f(0)=-3\). Therefore, by the intermediate value theorem, there is a \(z\) in \([-1,0]\) so that \(f(z)=0.\) Hence \(f\) has at least one real root.

The bisection method is an algorithm for finding solutions to an equation of the form \[f(x) = g(x),\] where \(f\) and \(g\) are continuous functions on an interval \([a,b]\).

Suppose that \(f(a) -g(a)\) is negative and that \(f(b)-g(b)\) is positive, and that \(c\) is the midpoint of \([a,b]\).

If \(f(c) - g(c)\) is negative, then there is a solution in \([c,b]\).

If \(f(c) - g(c)\) is positive, then there is a solution in \([a,c]\).

In either case, a solution is now guaranteed to exist in an interval that is half the length of the original interval.

Example 25

Use the bisection method to approximate a solution to the equation \[x^2 = 2\] to within an error of no greater than \(\tfrac{1}{20}\).

Take \(f(x)=x^2\) and \(g(x)=2\). Because they are polynomial functions, they are continuous everywhere. Focus on the interval \([1,2]\).

Notice that \(f(1)-g(1)=-1\) and \(f(2)-g(2)=2\) have opposite sign. Let \(c\) be the midpoint of the interval \([1,2]\): \[c=\frac{2+1}{2}=\frac{3}{2}.\]

The difference \(f(c)-g(c)=f\left(\tfrac{3}{2}\right)-g\left(\tfrac{3}{2}\right)=\tfrac{1}{4}\) is positive, so that means there is a solution in \(\left[1,\tfrac{3}{2}\right].\)

Let \(c\) be the midpoint of \(\left[1,\tfrac{3}{2}\right]\): \[c=\frac{\frac{3}{2}+1}{2}=\frac{5}{4}.\]

The difference \(f(c)-g(c)=f\left(\tfrac{5}{4}\right)-g\left(\tfrac{5}{4}\right)=-\tfrac{7}{16}\) is negative, so that means there is a solution in \(\left[\tfrac{5}{4},\tfrac{3}{2}\right].\)

Let \(c\) be the midpoint of \(\left[\tfrac{5}{4},\tfrac{3}{2}\right]\): \[c=\frac{\frac{5}{4}+\frac{3}{2}}{2}=\frac{11}{8}.\]

The difference \(f(c)-g(c)=f\left(\tfrac{11}{8}\right)-g\left(\tfrac{11}{8}\right)=-\tfrac{7}{64}\) is negative, so that means there is a solution in \(\left[\tfrac{11}{8},\tfrac{3}{2}\right].\)

Let \(c\) be the midpoint of \(\left[\tfrac{11}{8},\tfrac{3}{2}\right]\): \[c=\frac{\frac{11}{8}+\frac{3}{2}}{2}=\frac{23}{16}.\]

The difference \(f(c)-g(c)=f\left(\tfrac{23}{6}\right)-g\left(\tfrac{23}{6}\right)=\tfrac{17}{256}\approx 0.06640625\) is positive, so that means there is a solution in \(\left[\tfrac{11}{8},\tfrac{23}{16}\right].\)

Let \(c\) be the midpoint of \(\left[\tfrac{11}{8},\tfrac{23}{16}\right]\): \[c=\frac{\frac{11}{8}+\frac{23}{16}}{2}=\frac{45}{32}.\]

The difference \(f(c)-g(c)=f\left(\tfrac{45}{32}\right)-g\left(\tfrac{45}{32}\right)=-\tfrac{23}{1024}\approx -0.0224609375\) is negative, so that means there is a solution in \(\left[\tfrac{45}{32},\tfrac{23}{16}\right].\)

Let \(c\) be the midpoint of \(\left[\tfrac{45}{32},\tfrac{23}{16}\right]\): \[c=\frac{\frac{45}{32}+\frac{23}{16}}{2}=\frac{91}{64}.\]

The difference \(f(c)-g(c)=f\left(\tfrac{91}{64}\right)-g\left(\tfrac{91}{64}\right)=\tfrac{89}{4096}\approx 0.021728515625\) is positive, so that means there is a solution in \(\left[\tfrac{45}{32},\tfrac{91}{64}\right].\)

Therefore, the solution \(x^2=2\) is in the interval \([\tfrac{45}{32},\tfrac{91}{64}]\) and the error will be no greater than \(\tfrac{1}{20}=0.05\).

However, there is an exact solution to this equation: \(\sqrt{2}\). In fact, \(\sqrt{2}\) is in the interval \(\left[\tfrac{45}{32},\tfrac{91}{64}\right]\). This means \(\tfrac{45}{32}\leq\sqrt{2}\leq\tfrac{91}{64}\) or \(1.40625 \leq \sqrt{2}\leq 1.421875\). This is an approximation of the square root of two!

Return

• Return

© Copyright 2025 by the POC Writing Team: Bryan Carrillo, Yat Sun Poon, and David Weisbart. All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopy, recording, or any information storage and retrieval system, without permission in writing from the POC Writing Team.