Unit 1

Limits & Continuity

What a limit is, how to find one from a graph, a table or a formula, and what it means for a function to be continuous.

Topics in this unit

What a Limit Means · Limits from a Graph or Table · One-Sided Limits · Evaluating Limits · Squeeze Theorem · Infinite Limits & Asymptotes · Limits at Infinity · Continuity · Classifying & Removing Discontinuities · Intermediate Value Theorem

What a Limit Means

Question 1

no calculatormedium

For a function

f

, it is given that

lim_{x \to 3^{-}} f (x) = 4

lim_{x \to 3^{+}} f (x) = 4

, and

f (3) = 9

. Which statement must be true?

Show the answer

The answer is C.

The two one-sided limits agree (

4 = 4

), so the two-sided limit exists and equals

4

. The value

f (3) = 9

doesn't affect the limit; it only affects continuity (which fails here since

4 \neq = 9

Why not A:Continuity needs $lim_{x \to 3} f (x) = f (3)$ . Here $4 \neq = 9$ , so $f$ is discontinuous at $3$ .
Why not B: $9$ is the function value $f (3)$ , not the limit. The limit is determined by nearby behavior: $4$ .
Why not D:The limit DOES exist because both one-sided limits equal $4$ . The mismatch is with $f (3)$ , not between the sides.

Limits from a Graph or Table

Question 2

no calculatormedium

The graph of

f

is shown below. What is

x \to 2 lim f (x)

Show the answer

The answer is D.

Read each side of

x = 2

separately. From the left, the graph rises along the line to the open circle at

(2, 4)

, so

lim_{x \to 2^{-}} f (x) = 4

. From the right, the graph starts at the solid dot

(2, 1)

, so

lim_{x \to 2^{+}} f (x) = 1

. The two sides disagree, so

lim_{x \to 2} f (x)

does not exist. The hole at

x = - 1

and the corner at

x = 4

are the other features on the graph, and the limit exists at both of them.

Why not A: $1$ is $f (2)$ , the solid dot, and it is also the right-hand limit. From the left the graph climbs to $4$ , so the two sides disagree.
Why not B: $4$ is the left-hand limit, the open circle. From the right the graph starts down at $1$ , so the two sides disagree.
Why not C: $\frac{5}{2}$ averages the two one-sided limits. A limit is never an average: when the left side heads to $4$ and the right side to $1$ , there is no single value being approached.

Question 3

no calculatormedium

The graph of

f

is shown below. What is

x \to 1 lim f (x)

Show the answer

The answer is D.

From the left, the graph follows the line up to the solid dot

(1, 2)

, so

lim_{x \to 1^{-}} f (x) = 2

. From the right, the graph shoots upward along the vertical asymptote, so

lim_{x \to 1^{+}} f (x) = \infty

. One side settles at

2

while the other grows without bound, so the two-sided limit does not exist.

Why not A: $2$ is the left-hand limit and also $f (1)$ . The right side does not approach $2$ ; it climbs without bound.
Why not B: $\infty$ describes only the right side. To write $lim_{x \to 1} f (x) = \infty$ , both sides would have to grow without bound, and the left side stops at $2$ .
Why not C: $1$ is the horizontal asymptote, the height $f$ approaches as $x \to \infty$ , not as $x \to 1$ .

Question 4

no calculatorhard

The graph of the function

f

is shown below. The table gives values of the function

g

at selected values of

x

x g (x) 1.9 2.9 1.99 2.99 1.999 2.999 2.001 2.999 2.01 2.99 2.1 2.9

Assume the values of

g

continue the pattern shown in the table. What is

x \to 2 lim f (g (x))

Show the answer

The answer is B.

From the table, as

x \to 2

from either side,

g (x) \to 3

and every value is a little less than

3

. So the inputs to

f

approach

3

from the left, and

x \to 2 lim f (g (x)) = u \to 3^{-} lim f (u)

. On the graph, coming into

x = 3

from the left,

f

approaches the open circle at height

2

Why not A: $1$ is $f (3)$ , the solid dot. The inputs $g (x)$ get close to $3$ but never equal it, so the value at $3$ is never used.
Why not C: $4$ is the limit of $f$ from the right of $3$ . The table shows $g (x) < 3$ , so the inputs approach $3$ from the left.
Why not D: $u \to 3 lim f (u)$ does not exist, but that is not what is asked. Because $g (x)$ only approaches $3$ from below, only the left-hand limit of $f$ matters, and it exists.

Question 5

no calculatormedium

The graph of

f

is shown below. Which of the following limits does not exist?

Show the answer

The answer is C.

Check each point by asking whether the two sides head to the same height. At

x = - 2

there is a hole, but both sides approach

2

. At

x = 0

there is a corner, and both sides approach

0

. At

x = 4

the solid dot sits below the curve, but both sides approach

2

. At

x = 2

the left side approaches

2

while the right side starts at

4

. Only there do the sides disagree, so

lim_{x \to 2} f (x)

does not exist.

Why not A:A hole with no dot means $f (- 2)$ is undefined, but both sides still approach height $2$ , so the limit exists.
Why not B:A corner is not a break. Both sides come down to $(0, 0)$ , so the limit is $0$ .
Why not D:The solid dot at $(4, 0)$ is $f (4)$ , which the limit ignores. Both sides approach the open circle at height $2$ , so the limit is $2$ .

One-Sided Limits

Question 6

no calculatorhard

The graphs of the functions

f

and

g

are shown below. Which of the following limits exists?

Show the answer

The answer is C.

Work one side at a time. From the left,

f \to 2

and

g \to 1

. From the right,

f \to 0

and

g \to 3

. Then

f + g \to 2 + 1 = 3

from the left and

0 + 3 = 3

from the right, so

x \to 1 lim [f (x) + g (x)] = 3

. The two jumps cancel, so the sum has a limit even though neither

f

nor

g

does.

Why not A: $f$ jumps at $x = 1$ : it approaches $2$ from the left and $0$ from the right, so its limit does not exist.
Why not B: $g$ jumps at $x = 1$ : it approaches $1$ from the left and $3$ from the right, so its limit does not exist.
Why not D:The product approaches $2 \cdot 1 = 2$ from the left and $0 \cdot 3 = 0$ from the right. The sides disagree, so this limit does not exist.

Limits at Infinity

Question 7

no calculatormedium

The graph of

f

is shown below. What is

x \to - \infty lim f (x)

Show the answer

The answer is B.

x \to - \infty

(the far left), the graph levels off toward

y = - 1

. Functions can have different horizontal asymptotes in the two directions. That's perfectly fine, and the limit still exists in each direction separately. So

lim_{x \to - \infty} f (x) = - 1

Why not A: $1$ is the limit as $x \to + \infty$ , the right-hand end of the graph.
Why not C: $0$ is $f (0)$ , where the graph crosses the origin. The limit at $- \infty$ is about the far left, where the graph is flat at $- 1$ .
Why not D:A function can level off at different heights on the two ends. $lim_{x \to - \infty}$ and $lim_{x \to + \infty}$ are separate limits, and each one exists here.

Classifying & Removing Discontinuities

Question 8

no calculatormedium

The function

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

has a discontinuity at

x = 3

. Which statement best describes it?

Show the answer

The answer is C.

Factor:

\frac{( x - 3 ) ( x + 2 )}{x - 3} = x + 2

for

x \neq = 3

. The two-sided limit exists,

lim_{x \to 3} f (x) = 3 + 2 = 5

, but

f (3)

is undefined (a hole). A limit that exists at a point where the function is undefined is a removable discontinuity.

Why not A:The one-sided limits are equal (both $5$ ), so there's no jump. The factor $x - 3$ cancels.
Why not B:The $x - 3$ in the denominator cancels with the numerator, so there's no vertical asymptote, hence no infinite discontinuity.
Why not D: $f (3)$ is undefined (denominator $0$ ), so $f$ is not defined there. The discontinuity is real, just removable.

Intermediate Value Theorem

Question 9

no calculatormedium

The function

r

is continuous on

- 2 \leq x \leq 6

with

r (- 2) = 8

and

r (6) = - 4

. Which of the following must be true?

Show the answer

The answer is D.

Because

r

is continuous and changes sign from

r (- 2) = 8 > 0

r (6) = - 4 < 0

, the value

0

is intermediate, so the Intermediate Value Theorem guarantees a root: some

c

in the domain

(- 2, 6)

has

r (c) = 0

Why not A:The IVT guarantees only values between $r (6) = - 4$ and $r (- 2) = 8$ . Since $10 > 8$ , $r$ need not attain $10$ anywhere on $[- 2, 6]$ .
Why not B:Continuity does not force monotonicity; $r$ can wander up and down while still moving from $8$ to $- 4$ overall.
Why not C:The IVT guarantees each intermediate value is attained at least once, but offers no uniqueness; $r$ may hit a value several times.

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