how to find vertical asymptotes of tan

Learning Objectives

• ii.4.1 Explain the three conditions for continuity at a point.
• 2.four.2 Describe three kinds of discontinuities.
• ii.four.3 Define continuity on an interval.
• 2.4.four Land the theorem for limits of composite functions.
• 2.4.5 Provide an example of the intermediate value theorem.

Many functions have the property that their graphs can be traced with a pencil without lifting the pencil from the page. Such functions are chosen continuous. Other functions accept points at which a pause in the graph occurs, just satisfy this property over intervals contained in their domains. They are continuous on these intervals and are said to have a aperture at a indicate where a break occurs.

We begin our investigation of continuity past exploring what it means for a function to have continuity at a indicate. Intuitively, a role is continuous at a particular bespeak if there is no break in its graph at that betoken.

Continuity at a Bespeak

Earlier nosotros look at a formal definition of what information technology means for a function to be continuous at a point, permit's consider various functions that fail to meet our intuitive notion of what information technology means to be continuous at a point. We then create a list of atmospheric condition that preclude such failures.

Our beginning function of interest is shown in Effigy ii.32. We see that the graph of $f\left(x\right)$ has a pigsty at a. In fact, $f\left(a\right)$ is undefined. At the very to the lowest degree, for $f\left(x\right)$ to exist continuous at a, we demand the following status:

$$\text{i.}f\left(a\right)\text{is defined.}$$

Figure ii.32 The function $f\left(x\right)$ is non continuous at a because $f\left(a\right)$ is undefined.

Yet, as we encounter in Effigy 2.33, this condition lone is bereft to guarantee continuity at the betoken a. Although $f\left(a\right)$ is defined, the function has a gap at a. In this instance, the gap exists because $\underset{\mathrm{ten}\to a}{\text{lim}}f\left(x\right)$ does not exist. We must add another condition for continuity at a—namely,

$$\text{two.}\underset{x\to a}{\text{lim}}f\left(\mathrm{ten}\right)\text{exists.}$$

Figure 2.33 The role $f\left(x\right)$ is not continuous at a because $\underset{\mathrm{10}\to a}{\text{lim}}f\left(\mathrm{ten}\right)$ does not exist.

Nevertheless, as nosotros run into in Effigy two.34, these two conditions by themselves practice not guarantee continuity at a betoken. The function in this effigy satisfies both of our first two atmospheric condition, but is notwithstanding non continuous at a. We must add a 3rd condition to our listing:

$$\text{iii.}\underset{\mathrm{10}\to a}{\text{lim}}f\left(x\right)=f\left(a\right).$$

Figure 2.34 The function $f\left(\mathrm{10}\right)$ is not continuous at a because $\underset{x\to a}{\text{lim}}f\left(x\right)\ne f\left(a\right).$

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

Definition

A office $f\left(x\right)$ is continuous at a indicate a if and only if the following three conditions are satisfied:

1. $f\left(a\right)$ is defined
2. $\underset{\mathrm{10}\to a}{\text{lim}}f\left(x\right)$ exists
3. $\underset{x\to a}{\text{lim}}f\left(\mathrm{10}\right)=f\left(a\right)$

A function is discontinuous at a point a if information technology fails to be continuous at a.

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

Problem-Solving Strategy

Problem-Solving Strategy: Determining Continuity at a Point

1. Check to see if $f\left(a\right)$ is defined. If $f\left(a\right)$ is undefined, we need get no farther. The role is non continuous at a. If $f\left(a\right)$ is defined, go along to step 2.
2. Compute $\underset{\mathrm{10}\to a}{\text{lim}}f\left(x\right).$ In some cases, nosotros may need to do this by first computing $\underset{\mathrm{ten}\to a^{-}}{\text{lim}}f\left(x\right)$ and $\underset{\mathrm{10}\to a^{+}}{\text{lim}}f\left(x\right).$ If $\underset{\mathrm{ten}\to a}{\text{lim}}f\left(\mathrm{10}\right)$ does not exist (that is, it is non a existent number), so the part is not continuous at a and the trouble is solved. If $\underset{x\to a}{\text{lim}}f\left(\mathrm{ten}\right)$ exists, so continue to footstep three.
3. Compare $f\left(a\right)$ and $\underset{\mathrm{ten}\to a}{\text{lim}}f\left(\mathrm{10}\right).$ If $\underset{x\to a}{\text{lim}}f\left(\mathrm{ten}\right)\ne f\left(a\right),$ then the function is not continuous at a. If $\underset{x\to a}{\text{lim}}f\left(x\right)=f\left(a\right),$ then the part is continuous at a.

The next three examples demonstrate how to utilise this definition to determine whether a part is continuous at a given point. These examples illustrate situations in which each of the weather condition for continuity in the definition succeed or fail.

Example 2.26

Determining Continuity at a Point, Status i

Using the definition, determine whether the function $f\left(\mathrm{10}\right)=(x^{\mathrm{two}}-\mathrm{four})\text{/}(x-2)$ is continuous at $x=2.$ Justify the conclusion.

Example ii.27

Determining Continuity at a Signal, Condition 2

Using the definition, decide whether the function $f\left(x\right)=\{\begin{array}{ll}-x^2+4& \text{if}x\le 3\hfill \\ 4x-\mathrm{viii}& \text{if}\mathrm{ten}>3\hfill \end{array}$ is continuous at $\mathrm{ten}=3.$ Justify the determination.

Instance two.28

Determining Continuity at a Point, Condition iii

Using the definition, determine whether the function $f\left(x\right)=\{\begin{array}{ll}\frac{\text{sin}x}{x}\hfill & \text{if}x\ne 0\hfill \\ \hfill \mathrm{one}& \text{if}x=0\hfill \end{array}$ is continuous at $x=0.$

Checkpoint ii.21

Using the definition, determine whether the role $f\left(\mathrm{10}\right)=\{\begin{array}{cc}2\mathrm{ten}+1\hfill & \text{if}x<1\hfill \\ \hfill \mathrm{two}& \text{if}x=1\hfill \\ -\mathrm{10}+4\hfill & \text{if}\mathrm{10}>1\hfill \end{array}$ is continuous at $x=1.$ If the part is not continuous at 1, indicate the condition for continuity at a point that fails to hold.

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

Theorem 2.viii

Continuity of Polynomials and Rational Functions

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

Proof

Previously, we showed that if $p\left(x\right)$ and $q\left(x\right)$ are polynomials, $\underset{\mathrm{ten}\to a}{\text{lim}}p\left(x\right)=p\left(a\right)$ for every polynomial $p\left(\mathrm{10}\right)$ and $\underset{x\to a}{\text{lim}}\frac{p\left(x\right)}{q\left(x\right)}=\frac{p\left(a\right)}{q\left(a\right)}$ every bit long as $q\left(a\right)\ne 0.$ Therefore, polynomials and rational functions are continuous on their domains.

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We at present use Continuity of Polynomials and Rational Functions to determine the points at which a given rational function is continuous.

Example 2.29

Continuity of a Rational Function

For what values of x is $f\left(x\right)=\frac{x+1}{\mathrm{10}-5}$ continuous?

Checkpoint ii.22

For what values of 10 is $f\left(x\right)=\mathrm{iii}{\mathrm{10}}^{\mathrm{four}}-4x^2$ continuous?

Types of Discontinuities

As we have seen in Example ii.26 and Example two.27, discontinuities have on several different appearances. We classify the types of discontinuities we have seen thus far as removable discontinuities, infinite discontinuities, or leap discontinuities. Intuitively, a removable discontinuity is a discontinuity for which at that place is a hole in the graph, a bound discontinuity is a noninfinite discontinuity for which the sections of the function do not meet upward, and an infinite discontinuity is a aperture located at a vertical asymptote. Figure ii.37 illustrates the differences in these types of discontinuities. Although these terms provide a handy way of describing iii common types of discontinuities, keep in mind that not all discontinuities fit neatly into these categories.

Effigy 2.37 Discontinuities are classified as (a) removable, (b) jump, or (c) infinite.

These three discontinuities are formally defined every bit follows:

Definition

If $f\left(x\right)$ is discontinuous at a, and so

1. $f$ has a removable discontinuity at a if $\underset{\mathrm{10}\to a}{\text{lim}}f\left(x\right)$ exists. (Notation: When nosotros state that $\underset{x\to a}{\text{lim}}f\left(x\right)$ exists, we mean that $\underset{x\to a}{\text{lim}}f\left(\mathrm{10}\right)=L,$ where L is a existent number.)
2. $f$ has a spring discontinuity at a if $\underset{\mathrm{ten}\to a^{-}}{\text{lim}}f\left(x\right)$ and $\underset{x\to a^{+}}{\text{lim}}f\left(\mathrm{ten}\right)$ both exist, just $\underset{x\to a^{-}}{\text{lim}}f\left(x\right)\ne \underset{x\to a^{+}}{\text{lim}}f\left(\mathrm{ten}\right).$ (Note: When we land that $\underset{x\to a^{-}}{\text{lim}}f\left(x\right)$ and $\underset{x\to a^{+}}{\text{lim}}f\left(x\right)$ both be, we mean that both are real-valued and that neither take on the values ±∞.)
3. $f$ has an infinite aperture at a if $\underset{x\to a^{-}}{\text{lim}}f\left(x\right)=\text{±}\infty$ and/or $\underset{\mathrm{ten}\to a^{+}}{\text{lim}}f\left(\mathrm{10}\right)=\text{±}\infty .$

Instance 2.xxx

Classifying a Aperture

In Case two.26, nosotros showed that $f\left(x\right)=\frac{x^2-4}{x-\mathrm{ii}}$ is discontinuous at $\mathrm{10}=\mathrm{two}.$ Allocate this discontinuity equally removable, jump, or infinite.

Example 2.31

Classifying a Aperture

In Case 2.27, nosotros showed that $f\left(\mathrm{10}\right)=\{\begin{array}{cc}-{\mathrm{10}}^2+\mathrm{iv}& \text{if}\mathrm{ten}\le 3\hfill \\ 4x-8& \text{if}\mathrm{ten}>\mathrm{iii}\hfill \end{array}$ is discontinuous at $x=3.$ Allocate this discontinuity equally removable, jump, or infinite.

Example 2.32

Classifying a Discontinuity

Determine whether $f\left(x\right)=\frac{x+2}{x+1}$ is continuous at −1. If the part is discontinuous at −i, classify the discontinuity as removable, jump, or infinite.

Checkpoint 2.23

For $f\left(x\right)=\{\begin{array}{cc}{\mathrm{10}}^2\hfill & \text{if}\mathrm{10}\ne 1\hfill \\ \hfill \mathrm{iii}& \text{if}x=1\hfill \end{array},$ decide whether f is continuous at one. If f is non continuous at 1, classify the discontinuity as removable, leap, or infinite.

Continuity over an Interval

At present that we accept explored the concept of continuity at a point, we extend that idea to continuity over an interval. As we develop this thought for different types of intervals, information technology 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 whatever ii points in the interval without lifting the pencil from the newspaper. In preparation for defining continuity on an interval, we begin by looking at the definition of what it means for a part to be continuous from the right at a point and continuous from the left at a bespeak.

Continuity from the Right and from the Left

A function $f\left(x\right)$ is said to be continuous from the right at a if $\underset{\mathrm{ten}\to a^{+}}{\text{lim}}f\left(x\right)=f\left(a\right).$

A office $f\left(x\right)$ is said to exist continuous from the left at a if $\underset{x\to a^{-}}{\text{lim}}f\left(x\right)=f\left(a\right).$

A function is continuous over an open interval if it is continuous at every point in the interval. A function $f\left(\mathrm{ten}\right)$ is continuous over a airtight interval of the grade $\left[a,b\right]$ if it is continuous at every betoken in $\left(a,b\right)$ and is continuous from the right at a and is continuous from the left at b. Analogously, a part $f\left(\mathrm{10}\right)$ is continuous over an interval of the grade $\left(a,b\right]$ if information technology is continuous over $\left(a,b\right)$ and is continuous from the left at b. Continuity over other types of intervals are defined in a similar fashion.

Requiring that $\underset{x\to a^{+}}{\text{lim}}f\left(\mathrm{ten}\right)=f\left(a\right)$ and $\underset{x\to b^{-}}{\text{lim}}f\left(\mathrm{ten}\right)=f\left(b\right)$ ensures that we tin trace the graph of the office from the betoken $\left(a,f\left(a\right)\right)$ to the point $\left(b,f\left(b\right)\right)$ without lifting the pencil. If, for example, $\underset{x\to a^{+}}{\text{lim}}f\left(x\right)\ne f\left(a\right),$ nosotros would need to lift our pencil to jump from $f\left(a\right)$ to the graph of the residue of the function over $\left(a,b\right].$

Example 2.33

Continuity on an Interval

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

Example 2.34

Continuity over an Interval

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

Checkpoint two.24

State the interval(s) over which the part $f\left(x\right)=\sqrt{\mathrm{ten}+3}$ is continuous.

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

Theorem 2.9

Composite Function Theorem

If $f\left(x\right)$ is continuous at Fifty and $\underset{x\to a}{\text{lim}}g\left(x\right)=L,$ then

$$\underset{x\to a}{\text{lim}}f\left(\mathrm{thou}\left(x\right)\right)=f\left(\underset{x\to a}{\text{lim}}\mathrm{one\; thousand}\left(x\right)\right)=f\left(L\right).$$

Before we move on to Example ii.35, retrieve that earlier, in the department on limit laws, we showed $\underset{x\to 0}{\text{lim}}\text{cos}x=1=\text{cos}\left(0\right).$ Consequently, we know that $f\left(\mathrm{10}\right)=\text{cos}x$ is continuous at 0. In Example two.35 we encounter how to combine this result with the composite function theorem.

Instance 2.35

Limit of a Blended Cosine Function

Evaluate $\underset{\mathrm{ten}\to \pi \text{/}2}{\text{lim}}\text{cos}\left(x-\frac{\pi }{\mathrm{two}}\right).$

Checkpoint 2.25

Evaluate $\underset{x\to \pi }{\text{lim}}\text{sin}\left(x-\pi \right).$

The proof of the next theorem uses the blended part theorem as well as the continuity of $f\left(x\right)=\text{sin}x$ and $g\left(x\right)=\text{cos}x$ at the point 0 to show that trigonometric functions are continuous over their entire domains.

Theorem 2.x

Continuity of Trigonometric Functions

Trigonometric functions are continuous over their unabridged domains.

Proof

We begin by demonstrating that $\text{cos}x$ is continuous at every real number. To practise this, nosotros must testify that $\underset{\mathrm{10}\to a}{\text{lim}}\text{cos}\mathrm{ten}=\text{cos}a$ for all values of a.

$\begin{array}{ccccc}\underset{\mathrm{ten}\to a}{\text{lim}}\text{cos}x\hfill & =\underset{x\to a}{\text{lim}}\text{cos}\left(\left(x-a\right)+a\right)\hfill & & & \text{rewrite}x=x-a+a\hfill \\ & =\underset{x\to a}{\text{lim}}\left(\text{cos}\left(x-a\right)\text{cos}a-\text{sin}\left(x-a\right)\text{sin}a\right)\hfill & & & \text{apply the identity for the cosine of the sum of two angles}\hfill \\ & =\text{cos}\left(\underset{\mathrm{10}\to a}{\text{lim}}\left(x-a\right)\right)\text{cos}a-\text{sin}\left(\underset{x\to a}{\text{lim}}\left(x-a\right)\right)\text{sin}a\hfill & & & \underset{\mathrm{10}\to a}{\text{lim}}\left(x-a\right)=0,\text{and}\text{sin}x\text{and}\text{cos}\mathrm{10}\text{are continuous at 0}\hfill \\ & =\text{cos}\left(0\right)\text{cos}a-\text{sin}\left(0\right)\text{sin}a\hfill & & & \text{evaluate cos(0) and sin(0) and simplify}\hfill \\ & =1·\text{cos}a-0·\text{sin}a=\text{cos}a.\hfill \end{array}$

The proof that $\text{sin}\mathrm{10}$ is continuous at every real number is coordinating. 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 constabulary.

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As you tin can see, the composite function theorem is invaluable in demonstrating the continuity of trigonometric functions. As nosotros go along our study of calculus, we revisit this theorem many times.

The Intermediate Value Theorem

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

Theorem 2.xi

The Intermediate Value Theorem

Let f be continuous over a closed, bounded interval $\left[a,b\right].$ If z is whatsoever real number between $f\left(a\right)$ and $f\left(b\right),$ then there is a number c in $\left[a,b\right]$ satisfying $f\left(c\right)=z$ in Effigy ii.38.

Figure two.38 There is a number $c\in \left[a,b\right]$ that satisfies $f\left(c\right)=z.$

Example 2.36

Awarding of the Intermediate Value Theorem

Evidence that $f\left(\mathrm{10}\right)=x-\text{cos}x$ has at least one zero.

Example ii.37

When Can You Apply the Intermediate Value Theorem?

If $f\left(x\right)$ is continuous over $\left[0,2\right],f\left(0\right)>0$ and $f\left(2\right)>0,$ can nosotros use the Intermediate Value Theorem to conclude that $f\left(x\right)$ has no zeros in the interval $[0,\mathrm{two}\text{]?}$ Explain.

Instance 2.38

When Can Yous Apply the Intermediate Value Theorem?

For $f\left(\mathrm{10}\right)=\mathrm{ane}\text{/}x,f\left(\mathrm{-1}\right)=\mathrm{-1}<0$ and $f\left(\mathrm{one}\right)=1>0.$ Can we conclude that $f\left(x\right)$ has a nix in the interval $\left[\mathrm{-1},\mathrm{ane}\right]?$

Checkpoint ii.26

Show that $f\left(x\right)=x^{\mathrm{iii}}-x^2-3x+1$ has a zero over the interval $\left[0,\mathrm{i}\right].$

Section ii.4 Exercises

For the following exercises, make up one's mind the bespeak(s), if any, at which each office is discontinuous. Classify any discontinuity as leap, removable, infinite, or other.

131.

$f\left(x\right)=\frac{\mathrm{one}}{\sqrt{x}}$

132 .

$f\left(\mathrm{ten}\right)=\frac{2}{{\mathrm{ten}}^2+\mathrm{i}}$

133.

$f\left(\mathrm{10}\right)=\frac{\mathrm{ten}}{{\mathrm{ten}}^2-x}$

134 .

$\mathrm{grand}\left(t\right)=t^{\mathrm{-1}}+1$

135.

$f\left(x\right)=\frac{5}{{\mathrm{due\; east}}^x-2}$

136 .

$f\left(x\right)=\frac{\left|x-2\right|}{x-\mathrm{ii}}$

137.

$H\left(x\right)=\text{tan}2\mathrm{10}$

138 .

$f\left(t\right)=\frac{t+\mathrm{three}}{t^{\mathrm{ii}}+5t+\mathrm{six}}$

For the post-obit exercises, decide if the function continuous at the given point. If information technology is discontinuous, what blazon of discontinuity is it?

139.

$f(\mathrm{ten})=\frac{2x^{\mathrm{ii}}-\mathrm{five}\mathrm{10}+3}{x-1}$ at $x=1$

140 .

$h\left(\theta \right)=\frac{\text{sin}\theta -\text{cos}\theta }{\text{tan}\theta }$ at $\theta =\pi$

141.

$g\left(u\right)=\{\begin{array}{ll}\frac{6u^{\mathrm{two}}+u-2}{\mathrm{ii}u-1}& \text{if}u\ne \frac{\mathrm{one}}{2}\hfill \\ \frac{\mathrm{vii}}{2}\hfill & \text{if}u=\frac{1}{2}\hfill \end{array},$ at $u=\frac{1}{2}$

142 .

$f\left(y\right)=\frac{\text{sin}\left(\pi y\right)}{\text{tan}\left(\pi y\right)},$ at $y=1$

143.

$f\left(\mathrm{ten}\right)=\{\begin{array}{ll}{\mathrm{ten}}^{\mathrm{two}}-e^{\mathrm{10}}& \text{if}x<0\hfill \\ \mathrm{10}-1& \text{if}\mathrm{ten}\ge 0\hfill \end{array},$ at $x=0$

144 .

$f\left(x\right)=\{\begin{array}{l}\mathrm{10}\text{sin}\left(x\right)\text{if}\mathrm{10}\le \pi \\ x\text{tan}\left(x\right)\text{if}x>\pi \end{array},$ at $x=\pi$

In the following exercises, detect the value(southward) of thousand that makes each function continuous over the given interval.

145.

$f\left(x\right)=\{\begin{array}{cc}\mathrm{three}x+2,\hfill & x<k\hfill \\ 2x-3,\hfill & k\le x\le 8\hfill \end{array}$

146 .

$f\left(\theta \right)=\{\begin{array}{ll}\hfill \text{sin}\theta ,& 0\le \theta <\frac{\pi }{2}\hfill \\ \text{cos}\left(\theta +k\right),\hfill & \frac{\pi }{2}\le \theta \le \pi \hfill \end{array}$

147.

$f\left(x\right)=\{\begin{array}{cc}\frac{x^2+\mathrm{iii}\mathrm{ten}+2}{x+2},\hfill & x\ne -2\hfill \\ \hfill k,& x=\mathrm{-2}\hfill \end{array}$

148 .

$f\left(x\right)=\{\begin{array}{cc}\hfill e^{\mathrm{chiliad}x},& 0\le \mathrm{10}<4\hfill \\ \mathrm{ten}+\mathrm{iii},\hfill & 4\le x\le \mathrm{viii}\hfill \end{array}$

149.

$f\left(\mathrm{ten}\right)=\{\begin{array}{cc}\hfill \sqrt{kx},& 0\le x\le 3\hfill \\ \mathrm{ten}+1,\hfill & 3<\mathrm{ten}\le \mathrm{x}\hfill \end{array}$

In the post-obit exercises, use the Intermediate Value Theorem (IVT).

150 .

Allow $h\left(x\right)=\{\begin{array}{ll}\mathrm{three}{x}^2-\mathrm{iv},& \mathrm{ten}\le 2\hfill \\ 5+\mathrm{four}\mathrm{ten},& x>2\hfill \end{array}$ Over the interval $\left[0,4\right],$ there is no value of ten such that $h\left(x\right)=10,$ although $h\left(0\right)<\mathrm{x}$ and $h\left(\mathrm{four}\right)>\mathrm{ten}.$ Explain why this does not contradict the IVT.

151.

A particle moving along a line has at each time t a position function $\mathrm{south}\left(t\right),$ which is continuous. Presume $s\left(2\right)=\mathrm{five}$ and $s\left(\mathrm{five}\right)=2.$ Some other particle moves such that its position is given past $h\left(t\right)=s\left(t\right)-t.$ Explain why there must be a value c for $\mathrm{ii}<c<\mathrm{v}$ such that $h\left(c\right)=0.$

152 .

[T] Use the statement "The cosine of t is equal to t cubed."

1. Write a mathematical equation of the argument.
2. Prove that the equation in part a. has at least 1 existent solution.
3. Utilize a calculator to find an interval of length 0.01 that contains a solution.

153.

Apply the IVT to determine whether $2^x=x^3$ has a solution in one of the intervals $\left[1.25,1.375\right]$ or $\left[1.375,\mathrm{ane.five}\right].$ Briefly explicate your response for each interval.

154 .

Consider the graph of the function $y=f\left(\mathrm{10}\right)$ shown in the following graph.

1. Find all values for which the role is discontinuous.
2. For each value in part a., state why the formal definition of continuity does non apply.
3. Classify each discontinuity as either spring, removable, or infinite.

155.

Allow $f\left(x\right)=\{\begin{array}{c}\mathrm{iii}x,\mathrm{10}>1\\ {\mathrm{10}}^3,x<\mathrm{one}\end{array}.$

1. Sketch the graph of f.
2. Is it possible to find a value k such that $f\left(\mathrm{ane}\right)=k,$ which makes $f\left(x\right)$ continuous for all real numbers? Briefly explain.

156 .

Let $f\left(x\right)=\frac{x^{\mathrm{four}}-1}{x^2-1}$ for $x\ne -1,1.$

1. Sketch the graph of f.
2. Is it possible to find values $k_{\mathrm{ane}}$ and $k_2$ such that $f\left(\mathrm{-1}\right)=k_1$ and $f\left(1\right)=k_2,$ and that makes $f\left(\mathrm{10}\right)$ continuous for all real numbers? Briefly explain.

157.

Sketch the graph of the part $y=f\left(\mathrm{10}\right)$ with backdrop i. through vi.

1. The domain of f is $\left(\text{−}\infty ,\text{+}\infty \right).$
2. f has an infinite aperture at $\mathrm{10}=\mathrm{-vi}.$
3. $f\left(\mathrm{-6}\right)=3$
4. $\underset{\mathrm{10}\to {\mathrm{-three}}^{-}}{\text{lim}}f\left(x\right)=\underset{\mathrm{ten}\to {\mathrm{-3}}^{+}}{\text{lim}}f\left(x\right)=2$
5. $f\left(\mathrm{-3}\right)=3$
6. f is left continuous but non right continuous at $x=\mathrm{iii}.$
7. $\underset{x\to -\infty }{\text{lim}}f\left(\mathrm{10}\right)=\text{−}\infty$ and $\underset{x\to +\infty }{\text{lim}}f\left(x\right)=\text{+}\infty$

158 .

Sketch the graph of the function $y=f\left(x\right)$ with properties i. through iv.

1. The domain of f is $\left[0,5\right].$
2. $\underset{x\to {\mathrm{i}}^{+}}{\text{lim}}f\left(\mathrm{10}\right)$ and $\underset{x\to 1^{-}}{\text{lim}}f\left(x\right)$ exist and are equal.
3. $f\left(\mathrm{10}\right)$ is left continuous but not continuous at $x=2,$ and correct continuous but non continuous at $x=3.$
4. $f\left(x\right)$ has a removable discontinuity at $\mathrm{ten}=1,$ a jump discontinuity at $x=2,$ and the following limits concur: $\underset{x\to 3^{-}}{\text{lim}}f\left(\mathrm{ten}\right)=\text{−}\infty$ and $\underset{\mathrm{10}\to {\mathrm{iii}}^{+}}{\text{lim}}f\left(x\right)=2.$

In the post-obit exercises, suppose $y=f\left(x\right)$ is defined for all x. For each description, sketch a graph with the indicated property.

159.

Discontinuous at $x=1$ with $\underset{x\to \mathrm{-1}}{\text{lim}}f\left(x\right)=\mathrm{-ane}$ and $\underset{x\to \mathrm{two}}{\text{lim}}f\left(\mathrm{10}\right)=\mathrm{four}$

160 .

Discontinuous at $x=2$ merely continuous elsewhere with $\underset{x\to 0}{\text{lim}}f\left(\mathrm{ten}\right)=\frac{\mathrm{ane}}{2}$

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

161.

$f\left(t\right)=\frac{2}{e^t-{\mathrm{east}}^{-t}}$ is continuous everywhere.

162 .

If the left- and correct-hand limits of $f\left(x\right)$ as $x\to a$ exist and are equal, and then f cannot exist discontinuous at $x=a.$

163.

If a part is not continuous at a point, and so it is non defined at that indicate.

164 .

Co-ordinate to the IVT, $\text{cos}x-\text{sin}x-x=2$ has a solution over the interval $\left[\mathrm{-one},1\right].$

165.

If $f\left(x\right)$ is continuous such that $f\left(a\right)$ and $f\left(b\right)$ have opposite signs, so $f\left(x\right)=0$ has exactly one solution in $\left[a,b\right].$

166 .

The office $f\left(x\right)=\frac{x^{\mathrm{two}}-\mathrm{four}x+\mathrm{iii}}{{\mathrm{10}}^2-1}$ is continuous over the interval $\left[0,3\right].$

167.

If $f\left(x\right)$ is continuous everywhere and $f\left(a\right),f\left(b\right)>0,$ then there is no root of $f\left(\mathrm{ten}\right)$ in the interval $\left[a,b\right].$

[T] The following problems consider the scalar course of Coulomb'southward law, which describes the electrostatic force between ii point charges, such as electrons. It is given by the equation $F\left(r\right)={\mathrm{yard}}_{\mathrm{eastward}}\frac{\left|q_1{q}_2\right|}{r^{\mathrm{ii}}},$ where $k_{\mathrm{due\; east}}$ is Coulomb'south constant, $q_i$ are the magnitudes of the charges of the two particles, and r is the altitude between the two particles.

168 .

To simplify the adding of a model with many interacting particles, after some threshold value $r=R,$ nosotros guess F every bit goose egg.

1. Explain the concrete reasoning backside this assumption.
2. What is the forcefulness equation?
3. Evaluate the force F using both Coulomb's law and our approximation, bold two protons with a charge magnitude of $1.6022\times {10}^{\mathrm{-19}}\text{coulombs (C)},$ and the Coulomb constant $m_e=8.988\times {\mathrm{x}}^9{\text{Nm}}^{\mathrm{ii}}\text{/}{\text{C}}^{\mathrm{two}}$ are 1 yard apart. Also, assume $R<1\text{m}.$ How much inaccuracy does our approximation generate? Is our approximation reasonable?
4. Is in that location any finite value of R for which this organisation remains continuous at R?

169.

Instead of making the force 0 at R, instead we let the forcefulness be 10⁻²⁰ for $r\ge R.$ Assume 2 protons, which have a magnitude of charge $1.6022\times {\mathrm{x}}^{\mathrm{-19}}\text{C},$ and the Coulomb constant ${\mathrm{grand}}_e=8.988\times {10}^9{\text{Nm}}^{\mathrm{two}}\text{/}{\text{C}}^2.$ Is in that location a value R that tin make this system continuous? If so, find it.

Recall the discussion on spacecraft from the affiliate opener. The following problems consider a rocket launch from Earth's surface. The force of gravity on the rocket is given by $F\left(d\right)=-kk\text{/}{d}^2,$ where m is the mass of the rocket, d is the distance of the rocket from the center of World, and k is a constant.

170 .

[T] Make up one's mind the value and units of k given that the mass of the rocket is 3 million kg. (Hint: The altitude from the center of World to its surface is 6378 km.)

171.

[T] After a certain distance D has passed, the gravitational effect of World becomes quite negligible, then we tin judge the forcefulness function by $F\left(d\right)=\{\begin{array}{ll}-\frac{mk}{d^2}\hfill & \text{if}d<D\hfill \\ \mathrm{ten,000}\hfill & \text{if}d\ge D\hfill \end{array}.$ Using the value of k plant in the previous practise, find the necessary condition D such that the forcefulness office remains continuous.

172 .

As the rocket travels away from Globe's surface, there is a distance D where the rocket sheds some of its mass, since it no longer needs the excess fuel storage. Nosotros can write this function as $F\left(d\right)=\{\begin{array}{l}-\frac{m_{\mathrm{one}}\mathrm{thou}}{d^2}\text{if}d<D\hfill \\ -\frac{m_{\mathrm{two}}k}{d^2}\text{if}d\ge D\hfill \end{array}.$ Is there a D value such that this function is continuous, assuming $m_{\mathrm{one}}\ne m_2?$

Prove the following functions are continuous everywhere

173.

$f\left(\theta \right)=\text{sin}\theta$

174 .

$g\left(\mathrm{ten}\right)=\left|\mathrm{ten}\right|$

175.

Where is $f\left(\mathrm{ten}\right)=\{\begin{array}{l}0\text{if}\mathrm{10}\text{is irrational}\\ 1\text{if}x\text{is rational}\end{array}$ continuous?

Source: https://openstax.org/books/calculus-volume-1/pages/2-4-continuity

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