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Mathematics LibreTexts

12.E: Introduction to Calculus (Exercises)

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    7390
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    12.1: Finding Limits - Numerical and Graphical Approaches

    In this section, we will examine numerical and graphical approaches to identifying limits.

    Verbal

    Explain the difference between a value at \(x=a\) and the limit as \(x\) approaches \(a\).

    The value of the function, the output, at \(x=a\) is \(f(a)\). When the \(\lim \limits_{x \to a}f(x)\) is taken, the values of \(x\) get infinitely close to \(a\) but never equal \(a\). As the values of \(x\) approach \(a\) from the left and right, the limit is the value that the function is approaching.

    Explain why we say a function does not have a limit as \(x\) approaches \(a\) if, as \(x\) approaches \(a\), the left-hand limit is not equal to the right-hand limit.

    Graphical

    For the following exercises, estimate the functional values and the limits from the graph of the function \(f\) provided in Figure.

    CNX_Precalc_Figure_12_01_201.jpg 

    \(\lim \limits_{x \to −2^−} f(x)\)

    –4

    \(\lim \limits_{x \to −2^+ }f(x)\)

    \(\lim \limits_{x \to −2 f(x)}\)

    –4

    \(f(−2)\)

    \(\lim \limits_{x \to −1^− f(x)}\)

    2

    \(\lim \limits_{x \to 1^+} f(x)\)

    \(\lim \limits_{x \to 1} f(x)\)

    does not exist

    \(f(1)\)

    \(\lim \limits_{x \to 4^−} f(x)\)

    4

    \(\lim \limits_{x \to 4^+} f(x)\)

    \(\lim \limits_{x \to 4} f(x)\)

    does not exist

    \(f(4)\)

    For the following exercises, draw the graph of a function from the functional values and limits provided.

    \(\lim \limits_{x \to 0^−} f(x)=2, \lim \limits_{x \to 0^+} f(x)=–3, \lim \limits_{x \to 2} f(x)=2, f(0)=4, f(2)=–1, f(–3) \text{ does not exist.}\)

    Answers will vary.

    \(\lim \limits_{x \to 2^−} f(x)=0,\lim \limits_{x \to 2^+} =–2,\lim \limits_{x \to 0} f(x)=3, f(2)=5, f(0)\)

    Answers will vary.

    \(\lim \limits_{ x \to 2^−}  f(x)=2,  \lim \limits_{ x \to 2^+} f(x)=−3, \lim \limits_{x \to 0} f(x)=5, f(0)=1, f(1)=0\)

    Answers will vary.

    \(\lim \limits_{x \to 3^−} f(x)=0, \lim \limits_{x \to 3^+} f(x)=5, \lim \limits_{x \to 5} f(x)=0, f(5)=4, f(3) \text{ does not exist.}\)

    Answers will vary.

    \( \lim \limits_{ x \to 4} f(x)=6,  \lim \limits_{ x \to 6^+} f(x)=−1, \lim \limits_{ x \to 0} f(x)=5, f(4)=6, f(2)=6\)

    Answers will vary.

    \( \lim \limits_{ x \to −3} f(x)=2, \lim \limits_{ x \to 1^+} f(x)=−2, \lim \limits_{ x \to 3} f(x)=–4, f(–3)=0, f(0)=0\)

    Answers will vary.

    \( \lim \limits_{ x \to π} f(x)=π^2, \lim \limits_{ x \to –π} f(x)=\frac{π}{2}, \lim \limits_{ x \to 1^-} f(x)=0, f(π)=\sqrt{2}, f(0) \text{ does not exist}.\)

    Answers will vary.

    For the following exercises, use a graphing calculator to determine the limit to 5 decimal places as \(x\) approaches 0.

    \(f(x)=(1+x)^{\frac{1}{x}}\)

    \(g(x)=(1+x)^{\frac{2}{x}}\)

    7.38906

    \(h(x)=(1+x)^{\frac{3}{x}}\)

    \(i(x)=(1+x)^{\frac{4}{x}}\)

    54.59815

    \(j(x)=(1+x)^{\frac{5}{x}}\)

    Based on the pattern you observed in the exercises above, make a conjecture as to the limit of \(f(x)=(1+x)^{\frac{6}{x}}, g(x)=(1+x)^{\frac{7}{x}},\) and \(h(x)=(1+x)^{\frac{n}{x}}.\)

    \(e^6≈403.428794,e^7≈1096.633158, e^n\)

    For the following exercises, use a graphing utility to find graphical evidence to determine the left- and right-hand limits of the function given as \(x\) approaches \(a\). If the function has a limit as \(x\) approaches \(a\),state it. If not, discuss why there is no limit.

    \((x)= \begin{cases} |x|−1, && \text{if }x≠1 \\ x^3, && \text{if }x=1 \end{cases} a=1 \)

    \((x)= \begin{cases} \frac{1}{x+1}, && \text{if } x=−2 \\ (x+1)^2, && \text{if } x≠−2 \end{cases} a=−2 \)

    \(\lim \limits_{x \to −2} f(x)=1\)

    Numeric

    For the following exercises, use numerical evidence to determine whether the limit exists at \(x=a\). If not, describe the behavior of the graph of the function near \(x=a\). Round answers to two decimal places.

    \(f(x)=\frac{x^2−4x}{16−x^2};a=4\)

    \(f(x)=\frac{x^2−x−6}{x^2−9};a=3\)

    \(\lim \limits_{x \to 3} (\frac{x^2−x−6}{x^2−9})=\frac{5}{6}≈0.83\)

    \(f(x)=\frac{x^2−6x−7}{x^2– 7x};a=7\)

    \(f(x)=\frac{x^2–1}{x^2–3x+2};a=1\)

    \(\lim \limits_{x \to 1}(\frac{x^2−1}{x^2−3x+2})=−2.00\)

    \(f(x)=\frac{1−x^2}{x^2−3x+2};a=1\)

    \(f(x)=\frac{10−10x^2}{x^2−3x+2};a=1\)

    \(\lim \limits_{x \to 1}(\frac{10−10x^2}{x^2−3x+2})=20.00\)

    \(f(x)=\frac{x}{6x^2−5x−6};a=\frac{3}{2}\)

    \(f(x)=\frac{x}{4x^2+4x+1};a=−\frac{1}{2}\)

    \(\lim \limits_{x \to \frac{−1}{2}}(\frac{x}{4x^2+4x+1})\) does not exist. Function values decrease without bound as \(x\) approaches –0.5 from either left or right.

    \(f(x)=\frac{2}{x−4}; a=4\)

    For the following exercises, use a calculator to estimate the limit by preparing a table of values. If there is no limit, describe the behavior of the function as \(x\) approaches the given value.

    \(\lim \limits_{x \to 0} \frac{7 \tan x}{3x}\)

    \(\lim \limits_{x \to 0} \frac{7 \tan x}{3x}=\frac{7}{3}\)

    CNX_Precalc_Figure_12_01_202.jpg

     

    \(\lim \limits_{x \to 4} \frac{x^2}{x−4}\)

    CNX_Precalc_Figure_12_01_203.jpg

     

    \(\lim \limits_{x \to 0}\frac{2 \sin x}{4 \tan x}\)

     

    \(\lim \limits_{x \to 0} \frac{2 \sin x}{4 \tan x}=\frac{1}{2}\)

    CNX_Precalc_Figure_12_01_204.jpg

    For the following exercises, use a graphing utility to find numerical or graphical evidence to determine the left and right-hand limits of the function given as \(x\) approaches \(a\). If the function has a limit as \(x\) approaches \(a\), state it. If not, discuss why there is no limit.

    \(\lim \limits_{x \to 0}e^{e^{\frac{1}{x}}}\)

    \(\lim \limits_{x \to 0}e^{e^{− \frac{1}{x^2}}}\)

    \(\lim \limits_{x \to 0}e^{e^{− \frac{1}{x^2}}}=1.0\)

    \(\lim \limits_{x \to 0} \frac{|x|}{x}\)

    \(\lim \limits_{x \to −1} \frac{|x+1|}{x+1}\)

    \(\lim \limits_{ x→−1^−}\frac{| x+1 |}{x+1}=\frac{−(x+1)}{(x+1)}=−1\) and  \(\lim \limits_{ x \to −1^+}\frac{| x+1 |}{x+1}=\frac{(x+1)}{(x+1)}=1\); since the right-hand limit does not equal the left-hand limit, \(\lim \limits_{ x to −1}\frac{|x+1|}{x+1}\) does not exist.

    \(\lim \limits_{ x \to 5} \frac{| x−5 |}{5−x}\)

     

    \(\lim \limits_{ x \to −1}\frac{1}{(x+1)^2}\)

    \(\lim \limits_{ x \to −1} \frac{1}{(x+1)^2}\) does not exist. The function increases without bound as \(x\) approaches \(−1\) from either side.

    \(\lim \limits_{ x \to 1} \frac{1}{(x−1)^3}\)

    \(\lim \limits_{ x \to 0} \frac{5}{1−e^{\frac{2}{x}}}\)

    \(\lim \limits_{ x \to 0} \frac{5}{1−e^{\frac{2}{x}}}\) does not exist. Function values approach 5 from the left and approach 0 from the right.

    Use numerical and graphical evidence to compare and contrast the limits of two functions whose formulas appear similar: \(f(x)=| \frac{1−x}{x} |\) and \(g(x)=| \frac{1+x}{x} |\) as \(x\) approaches 0. Use a graphing utility, if possible, to determine the left- and right-hand limits of the functions \(f(x)\) and \(g(x)\) as \(x\) approaches 0. If the functions have a limit as \(x\) approaches 0, state it. If not, discuss why there is no limit.

    Extensions

    According to the Theory of Relativity, the mass m m of a particle depends on its velocity \(v\). That is

    \[m=\dfrac{m_o}{\sqrt{1−(v^2/c^2)}}\]

    where \(m_o\) is the mass when the particle is at rest and \(c\) is the speed of light. Find the limit of the mass, \(m\), as \(v\) approaches \(c^−.\)

    Through examination of the postulates and an understanding of relativistic physics, as \(v→c, m→∞. \)Take this one step further to the solution,

    \[\lim \limits_{v \to c^−}m=\lim \limits_{v \to c^−} \dfrac{m_o}{\sqrt{1−(v^2/c^2)}}=∞\]

    Allow the speed of light, \(c\), to be equal to 1.0. If the mass, \(m\), is 1, what occurs to \(m\) as \(v \to c\)? Using the values listed in Table, make a conjecture as to what the mass is as \(v\) approaches 1.00.

     
    \(v\) \(m\)
    0.5 1.15
    0.9 2.29
    0.95 3.20
    0.99 7.09
    0.999 22.36
    0.99999 223.61

    12.2: Finding Limits - Properties of Limits

    Graphing a function or exploring a table of values to determine a limit can be cumbersome and time-consuming. When possible, it is more efficient to use the properties of limits, which is a collection of theorems for finding limits. Knowing the properties of limits allows us to compute limits directly.

    Verbal

    Give an example of a type of function \(f\) whose limit, as \(x\) approaches \(a,\) is \(f(a)\).

    If \(f\) is a polynomial function, the limit of a polynomial function as \(x\) approaches \(a\) will always be \(f(a)\).

    When direct substitution is used to evaluate the limit of a rational function as \(x\) approaches \(a\) and the result is \(f(a)=\frac{0}{0}\),does this mean that the limit of \(f\) does not exist?

    What does it mean to say the limit of \(f(x)\), as \(x\) approaches \(c\), is undefined?

    It could mean either (1) the values of the function increase or decrease without bound as \(x\) approaches \(c,\) or (2) the left and right-hand limits are not equal.

    Algebraic

    For the following exercises, evaluate the limits algebraically.

    \(\lim \limits_{x \to 0} (3)\)

    \(\lim \limits_{x \to 2}(\frac{−5x}{x^2−1})\)

    \(\frac{−10}{3}\)

    \(\lim \limits_{x \to 2}(\frac{x^2−5x+6}{x+2})\)

    \(\lim \limits_{x \to 3}(\frac{x^2−9}{x−3})\)

    6

    \(\lim \limits_{x \to −1}(\frac{x^2−2x−3}{x+1})\)

    \(\lim \limits_{x \to \frac{3}{2}}(\frac{6x^2−17x+12}{2x−3})\)

    \(\frac{1}{2}\)

    \(\lim \limits_{ x \to −\frac{7}{2}}(\frac{8x^2+18x−35}{2x+7})\)

    \(\lim \limits_{ x \to 3}(\frac{x^2−9}{x−5x+6})\)

    6

    \(\lim \limits_{ x \to −3} (\frac{−7x^4−21x^3}{−12x^4+108x^2})\)

    \(\lim \limits_{ x \to 3} (\frac{x^2+2x−3}{x−3})\)

    does not exist

    \(\lim \limits_{ h \to 0} (\frac{(3+h)^3−27}{h})\)

    \(\lim \limits_{ h \to 0} (\frac{(2−h)^3−8}{h})\)

    \(−12\)

    \(\lim \limits_{ h \to 0}(\frac{(h+3)^2−9}{h})\)

    \(\lim \limits_{ h \to 0} (\frac{\sqrt{5−h}−\sqrt{5}}{h})\)

    \(−\frac{\sqrt{5}}{10}\)

    \(\lim \limits_{ x \to 0} (\frac{\sqrt{3−x}−\sqrt{3}}{x})\)

    \(\lim \limits_{ x \to 9}(\frac{x^2−81}{3−x})\)

    \(−108\)

    \(\lim \limits_{ x \to 1}(\frac{\sqrt{x}−x^2}{1−\sqrt{x}})\)

    \(\lim \limits_{ x \to 0}(\sqrt{x1+2x}−1)\)

    1

    \(\lim \limits_{ x \to \frac{1}{2}}(\frac{x^2−\frac{1}{4}}{2x−1})\)

    \(\lim \limits_{ x \to 4} (\frac{x^3−64}{x^2−16})\)

    6

    \(\lim \limits_{ x \to 2^−} (\frac{|x−2|}{x−2})\)

    \(\lim \limits_{ x \to 2^+} (\frac{| x−2 |}{x−2})\)

    1

    \(\lim \limits_{ x \to 2}(\frac{| x−2 |}{x−2})\)

    \(\lim \limits_{ x \to 4^−}(\frac{| x−4 |}{4−x})\)

    1

    \(\lim \limits_{ x \to 4^+}(\frac{| x−4 |}{4−x})\)

    \(\lim \limits_{ x \to 4}(\frac{| x−4 |}{4−x})\)

    does not exist

    \(\lim \limits_{ x \to 2}(\frac{−8+6x−x^2}{x−2})\)

    For the following exercise, use the given information to evaluate the limits: \(\lim \limits_{x \to c}f(x)=3, \lim \limits_{x \to c} g(x)=5\)

    \(\lim \limits_{x \to c} [  2f(x)+\sqrt{g(x)}  ]\)

    \(6+\sqrt{5}\)

    \(\lim \limits_{x \to c} [  3f(x)+\sqrt{g(x)}  ]\)

    \(\lim \limits_{x \to c}\frac{f(x)}{g(x)}\)

    \(\frac{3}{5}\)

    For the following exercises, evaluate the following limits.

    \(\lim \limits_{x \to 2} \cos (πx)\)

    \(\lim \limits_{x \to 2} \sin (πx)\)

    0

    \(\lim \limits_{x \to 2} \sin (\frac{π}{x})\)

    \(f(x)= \begin{cases} 2x^2+2x+1, && x≤0 \\ x−3, && x>0 ; \end{cases} \lim \limits_{x \to 0^+}f(x)\)

    \(−3\)

    \(f(x)= \begin{cases} 2x^2+2x+1, && x≤0 \\  x−3, && x>0 ; \end{cases} \lim \limits_{x \to 0^−} f(x)\)

    \(f(x)= \begin{cases} 2x^2+2x+1, && x≤0 \\ x−3, && x>0 ; \end{cases} \lim \limits_{x \to 0}f(x)\)

    does not exist; right-hand limit is not the same as the left-hand limit.

    \(\lim \limits_{x \to 4} \frac{\sqrt{x+5}−3}{x−4}\)

    \(\lim \limits_{x \to 2^+} (2x−〚x〛)\)

    2

    \(\lim \limits_{x \to 2} \frac{\sqrt{x+7}−3}{x^2−x−2}\)

    \(\lim \limits_{x \to 3^+}\frac{x^2}{x^2−9}\)

    Limit does not exist; limit approaches infinity.

    For the following exercises, find the average rate of change\(\frac{f(x+h)−f(x)}{h}\).

    \(f(x)=x+1\)

    \(f(x)=2x^2−1\)

    \(4x+2h\)

    \(f(x)=x^2+3x+4\)

    \(f(x)=x^2+4x−100\)

    \(2x+h+4\)

    \(f(x)=3x^2+1\)

    \(f(x)= \cos (x)\)

    \(\frac{\cos (x+h)− \cos (x)}{h}\)

    \(f(x)=2x^3−4x\)

    \(f(x)=\frac{1}{x}\)

    \(\frac{−1}{x(x+h)}\)

    \(f(x)=\frac{1}{x^2}\)

    \(f(x)=\sqrt{x}\)

    \(\frac{−1}{\sqrt{x+h}+\sqrt{x}}\)

    Graphical

    Find an equation that could be represented by Figure.

    CNX_Precalc_Figure_12_02_201.jpg 

    Find an equation that could be represented by Figure.

    CNX_Precalc_Figure_12_02_202.jpg 

     

    \(f(x)=\frac{x^2+5x+6}{x+3}\)

    For the following exercises, refer to Figure.

    CNX_Precalc_Figure_12_02_203F.jpg 

    What is the right-hand limit of the function as \(x\) approaches 0?

    What is the left-hand limit of the function as \(x\) approaches 0?

    does not exist

    Real-World Applications

    The position function \(s(t)=−16t^2+144t\) gives the position of a projectile as a function of time. Find the average velocity (average rate of change) on the interval \([ 1,2 ]\).

    The height of a projectile is given by \(s(t)=−64t^2+192t\) Find the average rate of change of the height from \(t=1\) second to \(t=1.5\) seconds.

    52

    The amount of money in an account after \(t\) years compounded continuously at 4.25% interest is given by the formula \(A=A_0e^{0.0425t}\),where \(A_0\) is the initial amount invested. Find the average rate of change of the balance of the account from \(t=1\) year to \(t=2\) years if the initial amount invested is $1,000.00.

    12.3: Continuity

    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. We can check three different conditions to decide if a function is continuous at a particular number.

    Section Exercises

    Verbal

    State in your own words what it means for a function \(f\) to be continuous at \(x=c\).

    Informally, if a function is continuous at \(x=c\), then there is no break in the graph of the function at \(f(c)\), and \(f(c)\) is defined.

    State in your own words what it means for a function to be continuous on the interval \((a,b)\).

    Algebraic

    For the following exercises, determine why the function \(f\) is discontinuous at a given point \(a\) on the graph. State which condition fails.

     

    \[f(x)=ln | x+3 |,a=−3\]

    discontinuous at \(a=−3\); \(f(−3)\) does not exist

     

    \[f(x)= \ln | 5x−2 |,a=\dfrac{2}{5}\]

    \(f(x)=\frac{x^2−16}{x+4},a=−4\)

    removable discontinuity at  \(a=−4; f(−4)\) is not defined

    \(f(x)=\frac{x^2−16x}{x},a=0\)

    \(f(x)= \begin{cases} x,  &&  x≠3 \\ 2x, && x=3 \end{cases} a=3\)

    Discontinuous at \(a=3; \lim \limits_{x \to 3} f(x)=3,\) but \(f(3)=6,\) which is not equal to the limit.

    \(f(x) = \begin{cases} 5,  &&x≠0 \\ 3,  && x=0 \end{cases}  a=0\)

    \(f(x)= \begin{cases} \frac{1}{2−x}, && x≠2 \\ 3, &&x=2  \end{cases}  a=2\)

    \(\lim \limits_{x \to 2}f(x)\) does not exist.

    \(f(x)= \begin{cases} \frac{1}{x+6}, && x=−6 \\ x^2, && x≠−6 \end{cases}   a=−6\)

    \(f(x)=\begin{cases} 3+x, &&x<1 \\ x, &&x=1 \\ x^2, && x>1 \end{cases}    a=1\)

    \(\lim \limits_{x \to 1^−}f(x)=4;\lim \limits_{x \to 1^+}f(x)=1.\) Therefore, \(\lim \limits_{x \to 1}f(x)\) does not exist.

    \(f(x)= \begin{cases} 3−x, && x<1 \\ x, && x=1 \\ 2x^2, && x>1 \end{cases}    a=1\)

    \(f(x)= \begin{cases} 3+2x, && x<1 \\ x, && x=1 \\ −x^2, && x>1 \end{cases}    a=1\)

    \(\lim \limits_{x \to 1^−} f(x)=5≠ \lim \limits_{x \to 1^+}f(x)=−1\). Thus \(\lim \limits_{x \to 1}f(x)\) does not exist.

    \(f(x)= \begin{cases} x^2, &&x<−2 \\ 2x+1, && x=−2 \\ x^3, && x>−2  \end{cases}  a=−2\)

    \(f(x)= \begin{cases} \frac{x^2−9}{x+3}, && x<−3 \\ x−9, && x=−3 \\ \frac{1}{x}, && x>−3 \end{cases}    a=−3\)

    \(\lim \limits_{x to −3^+}f(x)=−\frac{1}{3}\) 

    Therefore, \(\lim \limits_{x \to −3} f(x)\) does not exist.

    \(f(x)= \begin{cases} \frac{x^2−9}{x+3}, && x<−3 \\ x−9, && x=−3\\ −6, && x>−3 \end{cases}   a=3\)

    \(f(x)=\frac{x^2−4}{x−2},  a=2\)

     \(f(2)\) is not defined.

    \(f(x)=\frac{25−x^2}{x^2−10x+25},  a=5\)

    \(f(x)=\frac{x^3−9x}{x^2+11x+24},  a=−3\)

    \(f(−3)\) is not defined.

    \(f(x)=\frac{x^3−27}{x^2−3x},  a=3\)

    \(f(x)=\frac{x}{|x|},  a=0\)

    \(f(0)\) is not defined.

    \(f(x)=\frac{2|x+2|}{x+2},  a=−2\)

    For the following exercises, determine whether or not the given function \(f\) is continuous everywhere. If it is continuous everywhere it is defined, state for what range it is continuous. If it is discontinuous, state where it is discontinuous.

    \(f(x)=x^3−2x−15\)

    Continuous on \((−∞,∞)\)

    \(f(x)=\frac{x^2−2x−15}{x−5}\)

    \(f(x)=2⋅3^{x+4}\)

    Continuous on \((−∞,∞)\)

    \(f(x)=− \sin (3x)\)

    \(f(x)=\frac{|x−2|}{x^2−2x}\)

    Discontinuous at \(x=0\) and\(x=2\)

    \(f(x)= \tan (x)+2\)

    \(f(x)=2x+\frac{5}{x}\)

    Discontinuous at \(x=0\)

    \(f(x)=\log _2 (x)\)

    \(f(x)= \ln x^2 \)

    Continuous on \((0,∞)\)

    \(f(x)=e^{2x}\)

    \(f(x)=\sqrt{x−4}\)

    Continuous on \([4,∞)\)

    \(f(x)= \sec (x)−3\)

    \(f(x)=x^2+ \sin (x)\)

    Continuous on \((−∞,∞)\).

    Determine the values of \(b\) and \(c\) such that the following function is continuous on the entire real number line.

     

    \[f(x)= \begin{cases}x+1, && 1<x<3 \\ x^2+bx+c, &&|x−2|≥1 \end{cases}\]

    Graphical

    For the following exercises, refer to Figure. Each square represents one square unit. For each value of \( a\), determine which of the three conditions of continuity are satisfied at \(x=a\) and which are not.

    CNX_Precalc_Figure_12_03_201F.jpg 

    \(x=−3\)

    1, but not 2 or 3

    \(x=2\)

    \(x=4\)

    1 and 2, but not 3

    For the following exercises, use a graphing utility to graph the function \(f(x)= \sin (\frac{12π}{x})\) as in Figure. Set the x-axis a short distance before and after 0 to illustrate the point of discontinuity.

    CNX_Precalc_Figure_12_03_202F.jpg 

    Which conditions for continuity fail at the point of discontinuity?

    Evaluate \(f(0)\).

    \(f(0)\) is undefined.

    Solve for \(x\) if \(f(x)=0\).

    What is the domain of \(f(x)\)?

    \((−∞,0)∪(0,∞)\)

    For the following exercises, consider the function shown in Figure.

    CNX_Precalc_Figure_12_03_203.jpg 

    At what x-coordinates is the function discontinuous?

    What condition of continuity is violated at these points?

    At \(x=−1\), the limit does not exist. At \(x=1, f(1)\) does not exist.

    At \(x=2\), there appears to be a vertical asymptote, and the limit does not exist.

    Consider the function shown in Figure. At what x-coordinates is the function discontinuous? What condition(s) of continuity were violated?

    CNX_Precalc_Figure_12_03_204.jpg 

    Construct a function that passes through the origin with a constant slope of 1, with removable discontinuities at \(x=−7\) and \(x=1\).

    \(\frac{x^3+6x^2−7x}{(x+7)(x−1)}\)

    The function \(f(x)=\frac{x^3−1}{x−1}\) is graphed in Figure. It appears to be continuous on the interval \([−3,3]\), but there is an x-value on that interval at which the function is discontinuous. Determine the value of \(x\) at which the function is discontinuous, and explain the pitfall of utilizing technology when considering continuity of a function by examining its graph.

    CNX_Precalc_Figure_12_03_205F.jpg 

    Find the limit limx→1f(x) limx→1f(x) and determine if the following function is continuous at \(x=1\):

    \[fx= \begin{cases} x^2+4 && x≠1 \\ 2 && x=1\end{cases}\]

    The function is discontinuous at \(x=1\) because the limit as \(x\) approaches 1 is 5 and \(f(1)=2\).

    The graph of \(f(x)= \frac{\sin (2x)}{x}\) is shown in Figure. Is the function \(f(x)\) continuous at \(x=0?\) Why or why not?

    CNX_Precalc_Figure_12_03_206.jpg 

    12.4: Derivatives

    Change divided by time is one example of a rate. The rates of change in the previous examples are each different. In other words, some changed faster than others. If we were to graph the functions, we could compare the rates by determining the slopes of the graphs.