Types of discontinuity

Learning goals

Distinguishes removable, jump, and infinite discontinuities.
Identifies the limit or function-value condition that fails at a discontinuity.
Finds the value that removes a removable discontinuity.

Difficulty: 1 – 5

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Types of discontinuity 1

Define $g(x)=\begin{cases}x^2,&x\ne-2\\1,&x=-2\end{cases}$. Classify the discontinuity at $x=-2$, and state the single value that would make the function continuous there.

Hints

- Determine the nearby limiting value from the rule used away from the point. - Compare that limit with the assigned point value. - Ask whether changing only one function value could restore continuity.

Solution

1. For nearby inputs, $g(x)=x^2$, so $\lim_{x\to-2}g(x)=4$. 2. The actual value is $g(-2)=1$, which does not equal the limit. 3. The discontinuity is removable. Redefining $g(-2)=4$ would make the function continuous at $-2$.

Answer

The discontinuity is removable, and the repairing value is $g(-2)=4$.

Types of discontinuity 2

At $x=-1$, is the discontinuity in the graph removable, a jump, or infinite? What single change would make the function continuous there?

Hints

- Follow the curve on either side of the open circle. - Compare its limiting height with the filled point. - Would changing just one function value fix the break?

Solution

1. The curved branch approaches height $1$ from both sides of $x=-1$, but its value there is $3$. 2. The two-sided limit exists and is finite, so the discontinuity is removable. 3. Changing only $f(-1)$ to $1$ would make it continuous.

Answer

Removable; set $f(-1)=1$.

Types of discontinuity 3

Classify the discontinuity of $f$ at $x=0$. Would changing only $f(0)$ remove it? Explain using the two sides of the graph.

Hints

- Read the height reached by each branch near zero. - What kind of discontinuity has unequal finite one-sided limits? - Think about whether moving one dot changes the surrounding branches.

Solution

1. The left-hand branch approaches $2$ as $x\to0^-$. 2. The right-hand branch approaches $4$ as $x\to0^+$. 3. The finite one-sided limits are different, so there is a jump discontinuity. Changing the value at one point cannot make those limits agree.

Answer

Jump discontinuity; no single change to $f(0)$ can remove it.

Types of discontinuity 4

Classify the discontinuity of $f$ at $x=1$. Explain what each branch does near that input, and state any vertical asymptote.

Hints

- Does either branch approach a finite height near the excluded input? - What type of discontinuity involves unbounded values? - Name the x-value of the vertical line between the branches.

Solution

1. From the left of $1$, $f(x)$ falls without bound. 2. From the right of $1$, $f(x)$ rises without bound. 3. This is an infinite discontinuity, and the vertical asymptote is $x=1$.

Answer

Infinite discontinuity; vertical asymptote $x=1$.

Types of discontinuity 5

Let $f(x)=\frac{x^2-1}{x-1}$. Evaluate this statement: “Because the rule simplifies to $f(x)=x+1$ for $x\ne1$, the function is continuous at $x=1$.”

Hints

- Check whether the original function has a value at $x=1$. - Distinguish a limit from an actual function value. - Canceling a factor does not restore an excluded input automatically.

Solution

1. The original denominator is $0$ at $x=1$, so $f(1)$ is not defined. 2. Factoring gives $\frac{(x-1)(x+1)}{x-1}=x+1$ only for $x\ne1$. 3. Although $\lim_{x\to1}f(x)=2$, continuity at $x=1$ requires the function to be defined there and to satisfy $f(1)=2$. 4. Therefore, the statement is false. The graph has a removable discontinuity, or hole, at $(1, 2)$.

Answer

The statement is false. The limit is $2$, but $f(1)$ is undefined, so $f$ is not continuous at $x=1$. The discontinuity is removable.

Types of discontinuity 6

A hypothetical vehicle registration fee charges $\$2.00$ for each started $100\,\text{cm}^3$ of engine displacement. Let $H(V)$ be the fee in dollars for displacement $V>0$, measured in cubic centimeters. a) Find $H(1200)$, $H(1201)$, and $H(1299)$. b) Describe the portion of the graph for $1100\le V\le1400$. Name the function type. Treating $H$ as the full function on $V>0$, list every $V$-value in $[1100,1400]$ at which $H$ has a jump discontinuity.

Hints

- “Started” means that any partial $100\,\text{cm}^3$ block counts as the next block. - Distinguish the displayed interval from the domain of the full function. - Check the behavior immediately to the left and right of each multiple of $100$.

Solution

1. A displacement of $1200\,\text{cm}^3$ uses exactly $12$ started blocks, so $H(1200)=12\cdot\$2.00=\$24.00$. 2. Both $1201\,\text{cm}^3$ and $1299\,\text{cm}^3$ start a thirteenth block, so each fee is $13\cdot\$2.00=\$26.00$. 3. The graph is a step function. It is constant between consecutive multiples of $100$ and jumps immediately after each multiple of $100$. 4. For the full function on $V>0$, the multiples $1100$, $1200$, $1300$, and $1400$ are all jump discontinuities, and all four lie in the requested interval.

Answer

a) $H(1200)=\$24.00$, $H(1201)=\$26.00$, and $H(1299)=\$26.00$ b) The graph is a step function. The jump discontinuities of the full function that lie in $[1100,1400]$ are $V=1100,1200,1300,1400$.

Types of discontinuity 7

Let $g(x)=\frac{x^2-4x+3}{x-3}$. a) Find the maximal domain. b) Describe the function values near the excluded input. c) Can $g$ be extended continuously at $x=3$? If so, state the value that must be assigned to $g(3)$.

Hints

- Find the input that makes the denominator $0$. - Factor and simplify for allowed inputs. - Use the simplified rule to find the missing limiting value.

Solution

1. The denominator is $0$ at $x=3$, so the domain is $\mathbb{R}\setminus\{3\}$. 2. Factor the numerator: $x^2-4x+3=(x-3)(x-1)$. For $x\ne3$, $g(x)=x-1$. 3. Therefore, $\lim_{x\to3}g(x)=\lim_{x\to3}(x-1)=2$. The function values approach $2$ from both sides. 4. The discontinuity is removable. Defining $g(3)=2$ makes the extended function continuous.

Answer

a) $\mathbb{R}\setminus\{3\}$ b) The function values approach $2$ as $x\to3$. c) Yes. Define $g(3)=2$.

Types of discontinuity 8

Analyze each function at the indicated discontinuity $x_0$. Evaluate the limit if it exists, and classify the discontinuity as removable or infinite. a) $f(x)=\frac{x^2-9}{2x-6}$ at $x_0=3$ b) $g(x)=\frac{x-1}{x^2-2x+1}$ at $x_0=1$

Hints

- What result do you get from direct substitution? - Factor the numerator or denominator to look for a common factor. - If the factor causing the discontinuity cancels completely, what type of discontinuity remains? - If an uncanceled denominator factor approaches zero, compare the behavior from the left and right.

Solution

1. For part a, direct substitution gives $\frac{0}{0}$. Factor: $f(x)=\frac{(x-3)(x+3)}{2(x-3)}$. 2. For $x\neq 3$, $f(x)=\frac{x+3}{2}$. Therefore, $\lim_{x\to 3}f(x)=\frac{3+3}{2}=3$. The finite limit shows that $x=3$ is a removable discontinuity. 3. For part b, direct substitution also gives $\frac{0}{0}$. Factor: $g(x)=\frac{x-1}{(x-1)^2}$. 4. For $x\neq 1$, $g(x)=\frac{1}{x-1}$. As $x\to 1^-$, $g(x)\to -\infty$, and as $x\to 1^+$, $g(x)\to +\infty$. 5. The two-sided limit does not exist, and $x=1$ is an infinite discontinuity.

Answer

a) $\lim_{x\to 3}f(x)=3$; the discontinuity is removable. b) $\lim_{x\to 1}g(x)$ does not exist because the one-sided limits are $-\infty$ and $+\infty$; the discontinuity is infinite.

Types of discontinuity 9

The graph of $g$ has a special feature at $x=2$. a) Use the graph to determine whether $g$ is continuous at $x=2$. b) What value would $g(2)$ need to have for the function to be continuous there?

Hints

- An open circle shows a value the graph approaches but does not include. - A filled point gives the actual function value. - Continuity requires the limit to equal the function value.

Solution

1. As $x$ approaches $2$ from either side, the graph approaches $3$. Therefore, $\lim_{x\to2}g(x)=3$. 2. The filled point gives the actual function value $g(2)=1$. 3. Since $\lim_{x\to2}g(x)\ne g(2)$, the function has a removable discontinuity at $x=2$. 4. To make the function continuous, the missing point on the line must be filled, so the function must be redefined with $g(2)=3$.

Answer

a) $g$ is not continuous at $x=2$ because $\lim_{x\to2}g(x)=3$, but $g(2)=1$. b) $g(2)=3$

Types of discontinuity 10

Let $f(x)=\begin{cases}x+2,&x<1\\7,&x=1\\5-x,&x>1\end{cases}$. Determine whether $f$ is continuous at $x=1$. If not, classify the discontinuity.

Hints

- Evaluate the left-hand and right-hand limits using different branches. - Compare the two one-sided limiting values before looking at the point value. - A finite mismatch between the sides has a specific discontinuity classification.

Solution

1. From the left, $f(x)=x+2$, so $\lim_{x\to1^-}f(x)=3$. 2. From the right, $f(x)=5-x$, so $\lim_{x\to1^+}f(x)=4$. 3. The one-sided limits are finite but unequal, so the two-sided limit does not exist. The discontinuity at $x=1$ is a jump discontinuity; the value $f(1)=7$ does not change that classification.

Answer

$f$ is not continuous at $x=1$. It has a jump discontinuity there.

Types of discontinuity 11

Let $h(x)=\begin{cases}3x-1,&x<2\\x^2+1,&x>2\end{cases}$, so $h(2)$ is undefined. Classify the discontinuity at $x=2$, and give the value that would produce a continuous extension.

Hints

- Evaluate each branch at the boundary value through a one-sided limit. - Check whether the two sides approach the same number. - If only the point value is missing, consider whether one assignment can repair continuity.

Solution

1. The left-hand limit is $3\cdot 2-1=5$. 2. The right-hand limit is $2^2+1=5$. 3. The two-sided limit exists and equals $5$, but the function is undefined at $2$. The discontinuity is removable, and defining $h(2)=5$ would remove it.

Answer

Removable discontinuity; define $h(2)=5$.

Types of discontinuity 12

Consider $p(x)=\frac{1}{(x-2)^2}$. Describe the behavior of $p(x)$ as $x\to2^-$ and as $x\to2^+$, then classify the discontinuity at $x=2$.

Hints

- Examine the sign of the squared denominator on both sides of $2$. - Determine what happens when a positive denominator approaches zero. - Match unbounded nearby behavior to its discontinuity type.

Solution

1. On either side of $2$, $(x-2)^2$ is positive and approaches $0$. 2. Therefore $p(x)\to+\infty$ from both sides. 3. The function has an infinite discontinuity at $x=2$, with vertical asymptote $x=2$.

Answer

$p(x)\to+\infty$ from both sides. The discontinuity is infinite, and $x=2$ is a vertical asymptote.

Types of discontinuity 13

Consider $q(x)=\ln((x+1)^2)$, with its real domain. Find the one-sided behavior as $x\to-1^-$ and $x\to-1^+$. Classify the discontinuity at $x=-1$.

Hints

- What positive value is placed inside the logarithm near $-1$? - How does the logarithm behave as its positive input approaches zero? - Connect unbounded behavior to the discontinuity type.

Solution

1. On both sides of $x=-1$, the expression $(x+1)^2$ is positive and approaches $0$. 2. Because $\ln u\to-\infty$ as $u\to0^+$, $q(x)\to-\infty$ from both sides. 3. The discontinuity is infinite, and $x=-1$ is a vertical asymptote.

Answer

$\lim_{x\to-1^-}q(x)=-\infty$ and $\lim_{x\to-1^+}q(x)=-\infty$. The discontinuity is infinite.

Types of discontinuity 14

Tomasz says $g(x)=\frac{\sin x}{x}$, with $x$ measured in radians, has an infinite discontinuity at $x=0$ because the denominator is zero there. Evaluate the claim and classify the discontinuity.

Hints

- Separate “undefined at the point” from “unbounded near the point.” - Determine the nearby limit before naming the discontinuity type. - Ask whether assigning one finite point value would repair continuity.

Solution

1. The function is undefined at $x=0$, but an excluded input does not by itself imply unbounded behavior. 2. The standard limit $\lim_{x\to0}\frac{\sin x}{x}=1$ is finite. 3. Therefore the discontinuity at $x=0$ is removable, not infinite. Defining $g(0)=1$ would make the function continuous there.

Answer

The claim is false. The discontinuity is removable, and the continuous extension uses $g(0)=1$.

Types of discontinuity 15

A table samples a function $h$ near $x=4$. <table><tr><th>$x$</th><td>$3.9$</td><td>$3.99$</td><td>$3.999$</td><td>$4.001$</td><td>$4.01$</td><td>$4.1$</td></tr><tr><th>$h(x)$</th><td>$-10$</td><td>$-100$</td><td>$-1000$</td><td>$1000$</td><td>$100$</td><td>$10$</td></tr></table> Use the table to classify the discontinuity suggested at $x=4$. Describe the one-sided behavior that supports the classification.

Hints

- Separate the table into inputs less than and greater than $4$. - Track both the sign and magnitude of the outputs on each side. - Match unbounded nearby behavior to the appropriate discontinuity type.

Solution

1. From the left of $4$, the outputs become negative with rapidly increasing magnitude, suggesting $h(x)\to-\infty$. 2. From the right, the outputs become positive with rapidly increasing magnitude, suggesting $h(x)\to+\infty$. 3. The unbounded one-sided behavior indicates an infinite discontinuity at $x=4$.

Answer

Infinite discontinuity at $x=4$; the table suggests $h(x)\to-\infty$ from the left and $h(x)\to+\infty$ from the right.

Types of discontinuity 16

Classify the discontinuity of $y=\tan x$ at $x=\frac{\pi}{2}$. State the left-hand and right-hand behavior that supports your classification.

Hints

- Rewrite tangent as a quotient of sine and cosine. - Track the sign of cosine on each side of $\frac{\pi}{2}$. - Unbounded one-sided behavior identifies the discontinuity type.

Solution

1. As $x\to\frac{\pi}{2}^-$, $\cos x\to0^+$ while $\sin x\to1$, so $\tan x\to+\infty$. 2. As $x\to\frac{\pi}{2}^+$, $\cos x\to0^-$ while $\sin x\to1$, so $\tan x\to-\infty$. 3. The function has an infinite discontinuity at $x=\frac{\pi}{2}$, which is a vertical asymptote.

Answer

Infinite discontinuity at $x=\frac{\pi}{2}$; the left-hand limit is $+\infty$ and the right-hand limit is $-\infty$.

Types of discontinuity 17

Consider $s(x)=\frac{x}{|x|}$ for $x\ne0$. Determine the one-sided limits at $x=0$ and classify the discontinuity.

Hints

- Rewrite $|x|$ separately for negative and positive inputs. - Find each one-sided constant value. - Classify the discontinuity from the relationship between the two finite one-sided limits.

Solution

1. For $x<0$, $|x|=-x$, so $s(x)=-1$. Thus $\lim_{x\to0^-}s(x)=-1$. 2. For $x>0$, $|x|=x$, so $s(x)=1$. Thus $\lim_{x\to0^+}s(x)=1$. 3. The one-sided limits are finite but unequal, so $x=0$ is a jump discontinuity.

Answer

$\lim_{x\to0^-}s(x)=-1$, $\lim_{x\to0^+}s(x)=1$; jump discontinuity at $x=0$.

Types of discontinuity 18

Define $r(x)=\begin{cases}\frac{1}{(x-1)^2},&x\ne1\\0,&x=1\end{cases}$. Mina says assigning $r(1)=0$ removes the discontinuity. Evaluate the claim and classify the discontinuity at $x=1$.

Hints

- Analyze nearby behavior before considering the assigned point value. - Ask whether the two-sided limit is finite. - A removable discontinuity can be repaired at one point; unbounded behavior cannot.

Solution

1. As $x\to1$ from either side, $(x-1)^2\to0^+$, so $\frac{1}{(x-1)^2}\to+\infty$. 2. The nearby behavior is unbounded, so no finite choice of $r(1)$ can make the function continuous at $1$. 3. The discontinuity is infinite, not removable.

Answer

The claim is false. The discontinuity at $x=1$ is infinite, and assigning a finite point value cannot remove it.

Types of discontinuity 19

A table gives values of $v$ near $x=-2$, and $v(-2)=6$. <table><tr><th>$x$</th><td>$-2.1$</td><td>$-2.01$</td><td>$-2.001$</td><td>$-1.999$</td><td>$-1.99$</td><td>$-1.9$</td></tr><tr><th>$v(x)$</th><td>$5.8$</td><td>$5.98$</td><td>$5.998$</td><td>$1.002$</td><td>$1.02$</td><td>$1.2$</td></tr></table> What type of discontinuity does the table suggest at $x=-2$? Why does $v(-2)$ matching the left-side values not ensure continuity? Explain what the finite table cannot prove.

Hints

- Compare the nearest values on each side of the input. - What must agree for continuity at a point? - Could a function behave differently between the listed inputs?

Solution

1. The values from the left move toward $6$, which equals $v(-2)$. 2. The values from the right move toward $1$. If these one-sided trends hold for all sufficiently close inputs, the limits differ and there is a jump discontinuity. 3. Continuity requires both one-sided limits and the point value to agree. A finite table suggests their behavior but cannot prove the limits or continuity classification.

Answer

The table suggests a jump discontinuity: left-side values approach $6$, while right-side values approach $1$. Matching the left trend is insufficient, and finite samples cannot prove the one-sided limits.

Types of discontinuity 20

Use the three graphs to classify the discontinuity at $x=0$ in each panel as removable, jump, or infinite. Justify each classification from the one-sided behavior and the plotted point value, when relevant.

Hints

- Compare the left-hand and right-hand behavior in each panel before considering any filled point. - A finite common limit with a mismatched point value is removable; unequal finite one-sided limits form a jump. - Unbounded behavior on either side indicates an infinite discontinuity.

Solution

1. In panel a), both sides approach $1$, while the filled point is at $(0,3)$. The discontinuity is removable. 2. In panel b), the left side approaches $1$ and the right side approaches $3$. The discontinuity is a jump. 3. In panel c), both branches increase without bound as $x\to0$. The discontinuity is infinite.

Answer

a) Removable b) Jump c) Infinite

Types of discontinuity 21

Consider $w(x)=\ln(x-2)$, whose real domain is $x>2$. Describe the behavior as $x\to2^+$ and classify the vertical-asymptote behavior at $x=2$.

Hints

- Use the domain to identify the available direction of approach. - Focus on the logarithm's input $x-2$. - Recall the logarithm's behavior as its input approaches $0$ from the positive side.

Solution

1. As $x\to2^+$, $x-2\to0^+$. 2. The natural logarithm decreases without bound as its positive input approaches $0$, so $w(x)\to-\infty$. 3. Thus $x=2$ is a vertical asymptote, corresponding to infinite discontinuity behavior at the domain boundary.

Answer

$w(x)\to-\infty$ as $x\to2^+$. The graph has vertical asymptote $x=2$, with infinite discontinuity behavior there.

Types of discontinuity 22

A function $g$ has $\lim_{x\to a^-}g(x)=-2$ and $\lim_{x\to a^+}g(x)=4$. You may redefine the single value $g(a)$ to any real number $c$. Is there a choice of $c$ that makes $g$ continuous at $a$? Classify the discontinuity.

Hints

- Decide whether a two-sided limit exists before choosing a point value. - Redefining one point changes only the function value, not nearby one-sided behavior. - A repair by point redefinition is possible only for one discontinuity type.

Solution

1. The left-hand and right-hand limits are finite but unequal. 2. Therefore the two-sided limit at $a$ does not exist. 3. Changing only $g(a)$ cannot change either one-sided limit, so no choice of $c$ makes the function continuous. The discontinuity is a jump.

Answer

No real value of $c$ can make $g$ continuous at $a$. The discontinuity is a jump.

Types of discontinuity 23

Define $F(x)=\begin{cases}\frac{x^2-4}{x-2},&x\ne2\\4,&x=2\end{cases}$. Aoife says $F$ still has a removable discontinuity at $x=2$ because the fraction is undefined there. Decide whether a discontinuity actually remains.

Hints

- Simplify the nearby formula while preserving the separate point definition. - Compare the nearby limit with the assigned value at $2$. - A function can be defined specifically to fill a hole that another formula would have had.

Solution

1. For $x\ne2$, $\frac{x^2-4}{x-2}=x+2$. 2. Therefore $\lim_{x\to2}F(x)=4$. 3. Since the defined value is also $F(2)=4$, the continuity condition is satisfied. The removable discontinuity has already been repaired, so no discontinuity remains at $x=2$.

Answer

No discontinuity remains at $x=2$; $F$ is continuous there.

Types of discontinuity 24

Two functions are undefined at $x=4$: $p(x)=\frac{x^2-2x-8}{x-4}$ and $q(x)=\frac{3}{x-4}$, for $x\ne4$. Which function can be made continuous by assigning a value at $x=4$? Give that value and explain why the other function cannot be fixed this way.

Hints

- Simplify the first quotient away from the missing point. - What do the one-sided values of the second function do? - Decide whether changing a single point can affect the nearby limit.

Solution

1. Factor $x^2-2x-8=(x-4)(x+2)$. For $x\ne4$, $p(x)=x+2$, so its limit at $4$ is $6$. Defining $p(4)=6$ repairs the hole. 2. The function $q(x)=3/(x-4)$ tends to $-\infty$ from the left and $+\infty$ from the right. 3. No value for $q(4)$ changes that unbounded behavior, so its infinite discontinuity is not removable.

Answer

Only $p$ can be repaired, by setting $p(4)=6$. The function $q$ has an infinite discontinuity.

Types of discontinuity 25

A hypothetical shipping company charges for packages according to weight $w$ in pounds: - Up to and including $2\,\text{lb}$: $\$4.95$ - More than $2\,\text{lb}$ and up to and including $5\,\text{lb}$: $\$6.95$ - More than $5\,\text{lb}$ and up to and including $10\,\text{lb}$: $\$10.45$ - More than $10\,\text{lb}$ and up to and including $31.5\,\text{lb}$: $\$18.45$ a) Find the shipping cost for packages weighing $1.8\,\text{lb}$, $5.0\,\text{lb}$, and $5.1\,\text{lb}$. b) Write the cost function $K(w)$ as a piecewise function. c) Determine whether $K$ is continuous at $w=5$. Justify your answer using the left-hand and right-hand limits.

Hints

- Pay attention to whether each boundary value is included in the lower or upper interval. - Use piecewise notation to show a different constant value on each weight interval. - Continuity requires the two one-sided limits to be equal.

Solution

1. The package weighing $1.8\,\text{lb}$ is in the first interval, so its cost is $\$4.95$. The package weighing $5.0\,\text{lb}$ is included in the second interval, so its cost is $\$6.95$. The package weighing $5.1\,\text{lb}$ is in the third interval, so its cost is $\$10.45$. 2. The cost function is $K(w)=\begin{cases}\$4.95&\text{if }0<w\le2\\\$6.95&\text{if }2<w\le5\\\$10.45&\text{if }5<w\le10\\\$18.45&\text{if }10<w\le31.5\end{cases}$. 3. At $w=5$, $\lim_{w\to5^-}K(w)=\$6.95$, while $\lim_{w\to5^+}K(w)=\$10.45$. 4. Since the one-sided limits are different, the two-sided limit does not exist. The function has a jump discontinuity at $w=5$.

Answer

a) $1.8\,\text{lb}:\ \$4.95$; $5.0\,\text{lb}:\ \$6.95$; $5.1\,\text{lb}:\ \$10.45$ b) $K(w)=\begin{cases}\$4.95&0<w\le2\\\$6.95&2<w\le5\\\$10.45&5<w\le10\\\$18.45&10<w\le31.5\end{cases}$ c) Not continuous, because $\lim_{w\to5^-}K(w)=\$6.95\ne\$10.45=\lim_{w\to5^+}K(w)$.

Types of discontinuity 26

Consider the rational function $f(x)=\frac{x^2-1}{x^2-3x+2}$. a) State the domain of $f$. b) Evaluate $\lim_{x\to 1}f(x)$ and determine whether $\lim_{x\to 2}f(x)$ exists. c) Classify the discontinuity at each excluded value as removable or infinite.

Hints

- Find the zeros of the denominator to determine the excluded values. - Factor the numerator and denominator before evaluating the limits. - What does a canceled factor imply about the corresponding discontinuity? - Compare the left-hand and right-hand behavior near an uncanceled denominator zero.

Solution

1. Factor the denominator: $x^2-3x+2=(x-1)(x-2)$. Therefore, $D_f=\mathbb{R}\setminus\{1, 2\}$. 2. Factor the numerator: $x^2-1=(x-1)(x+1)$. For $x\neq 1$, $f(x)=\frac{x+1}{x-2}$. 3. Evaluate the first limit: $\lim_{x\to 1}f(x)=\frac{1+1}{1-2}=-2$. Because the limit is finite, $x=1$ is a removable discontinuity. 4. Near $x=2$, the numerator of the simplified expression approaches $3$, while the denominator approaches $0$. Specifically, $\lim_{x\to 2^-}f(x)=-\infty$ and $\lim_{x\to 2^+}f(x)=+\infty$. 5. The one-sided limits are different, so $\lim_{x\to 2}f(x)$ does not exist. The discontinuity at $x=2$ is infinite.

Answer

a) $D_f=\mathbb{R}\setminus\{1, 2\}$ b) $\lim_{x\to 1}f(x)=-2$. The limit $\lim_{x\to 2}f(x)$ does not exist because the one-sided limits are $-\infty$ and $+\infty$. c) The discontinuity at $x=1$ is removable, and the discontinuity at $x=2$ is infinite.

Types of discontinuity 27

Let $f(x)=\frac{x^2-2x-8}{x^2-16}$. Analyze each limit and interpret it geometrically. 1. $\lim_{x\to0}f(x)$ 2. $\lim_{x\to4}f(x)$ 3. $\lim_{x\to-4}f(x)$

Hints

- Try direct substitution first. - Factor the numerator and denominator when you obtain $\frac{0}{0}$. - Use one-sided signs when a denominator approaches zero without a canceling factor.

Solution

1. Direct substitution gives $f(0)=\frac{-8}{-16}=\frac{1}{2}$. This is the y-intercept. 2. Factor: $f(x)=\frac{(x-4)(x+2)}{(x-4)(x+4)}$. For $x\ne4$, this simplifies to $\frac{x+2}{x+4}$. Therefore, the limit at $4$ is $\frac{6}{8}=\frac{3}{4}$. The original graph has a removable discontinuity, or hole, at $\left(4, \frac{3}{4}\right)$. 3. Near $x=-4$, the simplified numerator approaches $-2$, while the denominator approaches $0$. From the left, the quotient approaches $+\infty$; from the right, it approaches $-\infty$. Therefore, the two-sided limit does not exist, and $x=-4$ is a vertical asymptote.

Answer

1. $\frac{1}{2}$ 2. $\frac{3}{4}$; removable discontinuity at $\left(4, \frac{3}{4}\right)$ 3. The two-sided limit does not exist: the left-hand limit is $+\infty$, and the right-hand limit is $-\infty$. The vertical asymptote is $x=-4$.

Types of discontinuity 28

Consider the rational function $g(x)=\frac{x^2-4}{x^3-2x^2}$. a) Find the discontinuities of $g$. b) Use limits to describe the behavior at each discontinuity. c) Classify each discontinuity as removable or infinite, and give the equation of any vertical asymptote.

Hints

- Factor the denominator to find all excluded values. - Factor the numerator and check for common factors. - If the denominator has greater multiplicity than the numerator at a common zero, what remains after simplification? - Use the simplified expression to evaluate the behavior near each excluded value.

Solution

1. Factor the denominator: $x^3-2x^2=x^2(x-2)$. The discontinuities occur at $x=0$ and $x=2$. 2. Factor the numerator: $x^2-4=(x-2)(x+2)$. 3. At $x=2$, cancel the common factor for $x\neq 2$: $g(x)=\frac{x+2}{x^2}$. Then $\lim_{x\to 2}g(x)=\frac{2+2}{2^2}=1$. Thus, $x=2$ is a removable discontinuity. 4. At $x=0$, use the simplified expression $\frac{x+2}{x^2}$. The numerator approaches $2$, and $x^2\to 0^+$ from both sides. Therefore, $\lim_{x\to 0}g(x)=+\infty$. 5. Thus, $x=0$ is an infinite discontinuity, the graph does not change sign across it, and the vertical asymptote is $x=0$.

Answer

a) The discontinuities are $x=0$ and $x=2$. b) $\lim_{x\to 2}g(x)=1$, and $\lim_{x\to 0}g(x)=+\infty$. c) The discontinuity at $x=2$ is removable. The discontinuity at $x=0$ is infinite, with vertical asymptote $x=0$.

Types of discontinuity 29

The graph shows a function $f$ on $[-4, 4]$. a) Read $f(-2)$, $f(1)$, and $f(3)$ from the graph. b) At which inputs in $[-4, 4]$ is $f$ not continuous? Briefly justify each answer from the graph.

Hints

- A filled point gives the actual function value; an open circle is excluded. - Look for places where the graph has a hole or a jump. - Compare the values approached from the left and right at each break. - For a removable discontinuity, the limit exists but differs from the function value.

Solution

1. At $x=-2$, the filled point is at $(-2, 3)$, so $f(-2)=3$. The open circle at $(-2, 1)$ is not included. 2. At $x=1$, the filled point is at $(1, 0)$, so $f(1)=0$. The open circle at $(1, 2.5)$ is not included. 3. At $x=3$, the graph lies on the line segment at $y=-2$, so $f(3)=-2$. 4. At $x=-2$, both one-sided limits equal $1$, but $f(-2)=3$. Therefore, $f$ has a removable discontinuity there. 5. At $x=1$, the left-hand limit is $2.5$, while the right-hand limit is $0$. Therefore, $f$ has a jump discontinuity there. The graph is continuous at all other inputs in the interval.

Answer

a) $f(-2)=3$, $f(1)=0$, and $f(3)=-2$ b) $f$ is not continuous at $x=-2$ and $x=1$. At $x=-2$, the discontinuity is removable; at $x=1$, it is a jump discontinuity.

Types of discontinuity 30

The graph shows a piecewise-defined function $f$ with a jump discontinuity at $x=1$. A new function $h$ is created by shifting the graph of $f$ $2$ units to the right and then reflecting it across the x-axis. a) At what input $x_0$ does $h$ have a discontinuity? b) Find $\lim_{x\to x_0^-}h(x)$, $\lim_{x\to x_0^+}h(x)$, and $h(x_0)$.

Hints

- First determine how the horizontal shift changes the input of the discontinuity. - Reflection across the x-axis changes output values but not input values. - Read the original one-sided limits and function value at $x=1$. - Multiply the relevant output values by $-1$.

Solution

1. Shifting the graph $2$ units to the right moves the discontinuity from $x=1$ to $x_0=3$. Reflecting across the x-axis does not change its x-coordinate. 2. For the original function, $\lim_{x\to1^-}f(x)=0$, $\lim_{x\to1^+}f(x)=1$, and $f(1)=1$. 3. Reflection across the x-axis multiplies every output by $-1$. Therefore, $\lim_{x\to3^-}h(x)=0$, $\lim_{x\to3^+}h(x)=-1$, and $h(3)=-1$.

Answer

a) $x_0=3$ b) $\lim_{x\to3^-}h(x)=0$, $\lim_{x\to3^+}h(x)=-1$, and $h(3)=-1$

Types of discontinuity 31

Use the graph of $g$ to determine whether the function is continuous at $x=-1$ and at $x=2$. Explain each conclusion.

Hints

- Use filled points to identify actual function values. - Use open circles and nearby graph pieces to identify limiting values. - Compare the left-hand limit, right-hand limit, and function value at each input.

Solution

1. At $x=-1$, the graph approaches $1$ from both sides, so $\lim_{x\to-1}g(x)=1$. However, the filled point shows that $g(-1)=2$. Since the limit and function value are different, $g$ has a removable discontinuity at $x=-1$. 2. At $x=2$, the left-hand limit is $4$, the right-hand limit is $0$, and the filled point gives $g(2)=4$. Since the one-sided limits are different, $g$ has a jump discontinuity at $x=2$.

Answer

At $x=-1$, $g$ is not continuous; it has a removable discontinuity. At $x=2$, $g$ is not continuous; it has a jump discontinuity.

Types of discontinuity 32

The graph shows a function $f$. The function $g$ is formed by reflecting $f$ across the y-axis, so $g(x)=f(-x)$. a) State the input $x_f$ where $f$ has a jump discontinuity. b) State the input $x_g$ where $g$ has a discontinuity. c) Use one-sided limits to explain why $g$ is not continuous at $x_g$.

Hints

- Reflection across the y-axis changes the sign of each x-coordinate. - Locate the jump in the original graph first. - Reflection across the y-axis reverses the left and right sides of approach. - Compare the two one-sided limits at the reflected input.

Solution

1. The graph of $f$ has a jump at $x_f=1$: the left-hand limit is $0$, while the right-hand limit is $2$. 2. Reflection across the y-axis changes each input $x$ to $-x$, so the jump moves from $x=1$ to $x_g=-1$. 3. The reflection reverses the sides of approach. Thus $\lim_{x\to-1^-}g(x)=2$ and $\lim_{x\to-1^+}g(x)=0$. 4. The filled point on the original graph gives $f(1)=0$, so $g(-1)=f(1)=0$. Since the one-sided limits are different, $g$ is not continuous at $x=-1$.

Answer

a) $x_f=1$ b) $x_g=-1$ c) $\lim_{x\to-1^-}g(x)=2$ and $\lim_{x\to-1^+}g(x)=0$. Since the one-sided limits are different, $g$ has a jump discontinuity at $x=-1$.

Types of discontinuity 33

The graph of $f$ has a jump discontinuity at $x=1$. Define $k(x)=f(x-3)+1$. a) Describe how the graph of $k$ is obtained from the graph of $f$. b) At what x-value $x_0$ does $k$ have a jump discontinuity? c) Find $k(x_0)$, $\lim_{x\to x_0^-}k(x)$, and $\lim_{x\to x_0^+}k(x)$.

Hints

- Track how the horizontal translation moves the original jump. - Use the filled point for the function value. - Read the left and right branches separately for the one-sided limits. - Apply the vertical shift to all three y-values.

Solution

1. Replacing $x$ with $x-3$ shifts the graph right $3$ units, and adding $1$ shifts it up $1$ unit. 2. The jump at $x=1$ therefore moves to $x_0=1+3=4$. 3. The filled point on the original graph is $(1, -5)$, so $k(4)=f(1)+1=-5+1=-4$. 4. The left-hand limit of $f$ at $1$ is $-5$, so $\lim_{x\to4^-}k(x)=-5+1=-4$. 5. The right-hand limit of $f$ at $1$ is $-1$, so $\lim_{x\to4^+}k(x)=-1+1=0$.

Answer

a) Shift right $3$ units and up $1$ unit. b) $x_0=4$ c) $k(4)=-4$, $\lim_{x\to4^-}k(x)=-4$, and $\lim_{x\to4^+}k(x)=0$

Types of discontinuity 34

Let $f(x)=\begin{cases}x+1,&x<0\\5,&x=0\\1-x,&0<x<2\\x,&x\ge2\end{cases}$. Analyze the discontinuities at $x=0$ and $x=2$. Classify each one.

Hints

- Analyze each boundary independently. - At each boundary, compare the left-hand limit, right-hand limit, and point value. - A matching limit with the wrong point value differs fundamentally from unequal one-sided limits.

Solution

1. At $x=0$, the left-hand limit is $1$ and the right-hand limit is also $1$, but $f(0)=5$. Thus $x=0$ is a removable discontinuity. 2. At $x=2$, the left-hand limit from $1-x$ is $-1$. 3. From the right, the branch $x$ approaches $2$, so the one-sided limits differ. Thus $x=2$ is a jump discontinuity.

Answer

At $x=0$: removable discontinuity. At $x=2$: jump discontinuity.

Types of discontinuity 35

A function $f$ has a removable discontinuity at $x=-1$, with $\lim_{x\to-1}f(x)=4$ and $f(-1)=9$. Define $g(x)=2f(x+3)-5$. Find the input where $g$ is discontinuous, classify the discontinuity, and state the value that would make $g$ continuous there.

Hints

- First solve for the x-value that sends the transformed input to the original discontinuity. - Transform the original limiting value and point value separately. - Decide whether the transformed mismatch can still be fixed by changing one point value.

Solution

1. The input of $f$ equals $-1$ when $x+3=-1$, so the transformed discontinuity occurs at $x=-4$. 2. As $x\to-4$, $f(x+3)\to4$, so $g(x)\to2\cdot 4-5=3$. The actual value is $g(-4)=2\cdot 9-5=13$. 3. The discontinuity remains removable. Redefining $g(-4)=3$ would make $g$ continuous there.

Answer

The discontinuity is removable at $x=-4$, and the repairing value is $g(-4)=3$.

Types of discontinuity 36

Consider $u(x)=e^{1/x}$ for $x\ne0$. Analyze the one-sided behavior at $x=0$ and classify the discontinuity.

Hints

- Determine the behavior of the exponent $1/x$ separately from each side. - Translate those exponent limits through the exponential function. - A discontinuity can be classified as infinite even when only one side is unbounded.

Solution

1. As $x\to0^-$, $1/x\to-\infty$, so $e^{1/x}\to0$. 2. As $x\to0^+$, $1/x\to+\infty$, so $e^{1/x}\to+\infty$. 3. Because the function is unbounded on one side of $0$, it has an infinite discontinuity there.

Answer

$\lim_{x\to0^-}u(x)=0$, while $u(x)\to+\infty$ as $x\to0^+$. The discontinuity is infinite.

Types of discontinuity 37

For a real parameter $k$, define $f_k(x)=\begin{cases}x,&x<0\\2,&x=0\\x+k,&x>0\end{cases}$. Classify the discontinuity at $x=0$ for $k=0$ and for $k\ne0$.

Hints

- Find each one-sided limit in terms of $k$. - Separate the case where the one-sided limits agree from the case where they do not. - Only after the sides agree should you compare the common limit with the point value.

Solution

1. The left-hand limit is always $0$. 2. The right-hand limit is $k$. 3. If $k=0$, the two-sided limit is $0$ but $f_k(0)=2$, so the discontinuity is removable. If $k\ne0$, the finite one-sided limits differ, so the discontinuity is a jump.

Answer

If $k=0$: removable discontinuity. If $k\ne0$: jump discontinuity.

Types of discontinuity 38

Let $R(x)=\begin{cases}\frac{1}{x},&x<1,\ x\ne0\\0,&x=0\\x+2,&x\ge1\end{cases}$. Identify and classify the discontinuities at $x=0$ and $x=1$.

Hints

- Analyze the two target inputs separately. - At $0$, focus on unbounded behavior of the reciprocal branch. - At $1$, compare the finite boundary values approached by the two branches.

Solution

1. Near $x=0$, the nearby rule is $1/x$. It approaches $-\infty$ from the left and $+\infty$ from the right, so $x=0$ is an infinite discontinuity. The assigned value $R(0)=0$ does not change this. 2. At $x=1$, the left-hand limit from $1/x$ is $1$. 3. The right-hand limit from $x+2$ is $3$. Since these finite limits differ, $x=1$ is a jump discontinuity.

Answer

At $x=0$: infinite discontinuity. At $x=1$: jump discontinuity.

Types of discontinuity 39

For $F_a(x)=\frac{x^2-4}{x^2+(a-2)x-2a}$, analyze the possible discontinuities. a) Determine when $x=2$ is a removable discontinuity and when it is an infinite discontinuity. b) When the discontinuity at $x=2$ is removable, give the value that would make the continuous extension. c) For $a=1$, identify and classify every discontinuity and state each vertical asymptote.

Hints

- Examine the factors of the numerator and denominator. - What changes if both denominator factors vanish at the same input? - For the given parameter value, check every zero of the original denominator.

Solution

1. Factor $x^2-4=(x-2)(x+2)$ and $x^2+(a-2)x-2a=(x-2)(x+a)$. 2. If $a\ne-2$, cancel $x-2$ for nearby inputs. The limit at $2$ is $\frac{4}{a+2}$, making the original hole removable by defining $F_a(2)=\frac{4}{a+2}$. 3. If $a=-2$, the denominator is $(x-2)^2$ and the nearby expression is $\frac{x+2}{x-2}$; it is unbounded at $2$, so that point is an infinite discontinuity. 4. For $a=1$, the denominator is $(x-2)(x+1)$. The input $2$ is removable with limit $\frac{4}{3}$; the input $-1$ is an infinite discontinuity and vertical asymptote.

Answer

a) Removable when $a\ne-2$; infinite when $a=-2$. b) Set $F_a(2)=\frac{4}{a+2}$ when $a\ne-2$. c) For $a=1$, $x=2$ is removable, $x=-1$ is infinite, and $x=-1$ is the only vertical asymptote.

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Types of discontinuity

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