Calculus 1 Quiz: Estimating Limits From Tables

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Estimating Limits From TablesQuestion 1 of 20

The functions $f(x)$ , $g(x)$ , and $h(x)$ satisfy the inequality $g(x) \le f(x) \le h(x)$ for all $x$ near 5. The following are values for $g(x)$ and $h(x)$ . For $g(x)$ : $g(4.9)=-3.09$ , $g(4.99)=-3.009$ , $g(5.01)=-2.991$ , $g(5.1)=-2.91$ . For $h(x)$ : $h(4.9)=-2.905$ , $h(4.99)=-2.995$ , $h(5.01)=-3.005$ , $h(5.1)=-3.05$ . What is the best estimate for $\lim_{x \to 5} f(x)$ ?

-3

Cannot be determined.

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Calculus 1 Quiz

Calculus 1 Quiz: Estimating Limits From Tables

Practice Estimating Limits From Tables in Calculus 1 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Estimating Limits From Tables, giving you a quick way to practice the rules, question types, and explanations that matter most for Calculus 1.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

-3 (correct answer)
0
5
Cannot be determined.

Explanation: From the table for

g(x)

, we can estimate that

\lim_{x \to 5} g(x) = -3

. From the table for

h(x)

, we can estimate that

\lim_{x \to 5} h(x) = -3

. Since

g(x) \le f(x) \le h(x)

and both

g(x)

and

h(x)

approach the same limit of -3 as

x \to 5

, the Squeeze Theorem implies that

f(x)

must also approach -3. Therefore,

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

Question 2

A function $g(t)$ exhibits the following behavior near $t=0$ : $g(-0.1) = 0.09$ , $g(-0.01) = -0.009$ , $g(-0.001) = 0.0009$ , and $g(0.001) = -0.0009$ , $g(0.01) = -0.009$ , $g(0.1) = 0.09$ . What is the best estimate for $\lim_{t \to 0} g(t)$ ?

0 (correct answer)
0.09
-0.009
Does not exist.

Explanation: The values of

g(t)

are oscillating (alternating in sign) as

t

approaches 0. However, the magnitude of these values is decreasing and approaching 0. For example,

|g(-0.1)| = 0.09

|g(-0.01)| = 0.009

|g(-0.001)| = 0.0009

. As

t \to 0

, the values of

g(t)

are being 'squeezed' towards 0. Therefore, the limit is 0.

Question 3

The values of a function $P(t)$ as $t$ approaches 0 are recorded: $P(-0.1) = 2.7048$ , $P(-0.01) = 2.7181$ , $P(-0.001) = 2.71827$ , and $P(0.001) = 2.71829$ , $P(0.01) = 2.7184$ , $P(0.1) = 2.7320$ . The trend in this data suggests that $\lim_{t \to 0} P(t)$ is equal to which mathematical constant?

$\pi$
e (correct answer)
$\phi$ (the golden ratio)
3

Explanation: The values of

P(t)

are getting closer and closer to the value of the mathematical constant

e \approx 2.71828

. As

t

gets closer to 0, the function's output refines to more decimal places of

e

. This is strong evidence that the limit is

e

Question 4

A table of values for a function $f(x)$ near $x=3$ is given: $f(2.9)=8.8$ , $f(2.99)=8.98$ , $f(3.01)=9.02$ , $f(3.1)=9.2$ . Based on this data, estimate $\lim_{x \to 1} f(x+2)$ .

1
3
9 (correct answer)
Cannot be determined.

Explanation: We can use a change of variables. Let

u = x+2

. As

x \to 1

, the value of

u

approaches

1+2=3

. So, the problem is equivalent to finding

\lim_{u \to 3} f(u)

. The provided table shows the behavior of

f

as its input approaches 3. As the input gets closer to 3 from both sides, the output

f(u)

gets closer to 9. Therefore, the limit is 9.

Question 5

The velocity $v(t)$ in m/s of an object at time $t$ in seconds is recorded in the table: $v(1.99)=11.94$ , $v(1.999)=11.994$ , $v(2.001)=12.006$ , $v(2.01)=12.06$ . The instantaneous velocity at $t=2$ is defined as $\lim_{t \to 2} v(t)$ . What does the data suggest for the instantaneous velocity at $t=2$ seconds?

6 m/s
11.99 m/s
12 m/s (correct answer)
The acceleration at t=2.

Explanation: The question asks for the instantaneous velocity at

t=2

, which is the limit of

v(t)

t

approaches 2. Looking at the table, as

t

approaches 2 from the left (1.99, 1.999),

v(t)

approaches 12. As

t

approaches 2 from the right (2.01, 2.001),

v(t)

also approaches 12. Therefore, the best estimate for the limit, and thus the instantaneous velocity, is 12 m/s.

Question 6

Tables for $f(x)$ and $g(x)$ near $x=-2$ are given. For $f(x)$ : $f(-2.1)=-0.42$ , $f(-2.01)=-0.0402$ , $f(-1.99)=0.0398$ , $f(-1.9)=0.38$ . For $g(x)$ : $g(-2.1)=0.21$ , $g(-2.01)=0.0201$ , $g(-1.99)=-0.0199$ , $g(-1.9)=-0.19$ . Estimate $\lim_{x \to -2} \frac{f(x)}{g(x)}$ .

-2 (correct answer)
0
1
Does not exist.

Explanation: From the tables,

\lim_{x \to -2} f(x) = 0

and

\lim_{x \to -2} g(x) = 0

. This is an indeterminate form of type 0/0. To estimate the limit, we can examine the ratio

f(x)/g(x)

for values of

x

close to -2. At

x=-2.1

, the ratio is

-0.42/0.21 = -2

. At

x=-2.01

, the ratio is

-0.0402/0.0201 = -2

. At

x=-1.99

, the ratio is

0.0398/(-0.0199) = -2

. At

x=-1.9

, the ratio is

0.38/(-0.19) = -2

. Since the ratio consistently equals -2, the best estimate for the limit is -2.

Question 7

A table for a function $f(x)$ is given: $f(2.9) = 10.5$ , $f(2.99) = 10.95$ , $f(3) = 11$ , $f(3.01) = 11.05$ , $f(3.1) = 11.5$ . Using this data, estimate the value of $\lim_{h \to 0^+} f(3-h)$ .

10.95
11 (correct answer)
11.05
Does not exist.

Explanation: Let

x = 3-h

. As

h

approaches 0 from the positive side (

h \to 0^+

h

takes on small positive values. This means

x = 3-h

will be slightly less than 3. So, the expression

\lim_{h \to 0^+} f(3-h)

is equivalent to the left-hand limit

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

. Looking at the table for values of

x

approaching 3 from the left (2.9, 2.99), the function values (10.5, 10.95) are approaching 11. Thus, the limit is 11.

Question 8

As $x$ approaches 2 from the left, a function $f(x)$ takes the sequence of values $4.5, 4.9, 4.99, 4.999, \dots$ . As $x$ approaches 2 from the right, $f(x)$ takes the sequence of values $5.5, 5.1, 5.01, 5.001, \dots$ . What is the best estimate for $\lim_{x \to 2} (f(x)-5)^2$ ?

0 (correct answer)
0.25
1
Does not exist.

Explanation: First, we estimate

\lim_{x \to 2} f(x)

. As

x

approaches 2 from the left, the values of

f(x)

approach 5. As

x

approaches 2 from the right, the values of

f(x)

also approach 5. Therefore,

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

. Now we can evaluate the desired limit using limit laws:

\lim_{x \to 2} (f(x)-5)^2 = (\lim_{x \to 2} f(x) - 5)^2 = (5-5)^2 = 0^2 = 0

Question 9

Selected values for functions $f(x)$ and $g(x)$ are given. For $f(x)$ : $f(-0.1) = 4.9$ , $f(-0.01) = 4.99$ , $f(0.01) = 5.01$ , $f(0.1) = 5.1$ . For $g(x)$ : $g(-0.1) = -1.8$ , $g(-0.01) = -1.98$ , $g(0.01) = -2.02$ , $g(0.1) = -2.2$ . What is the best estimate for $\lim_{x \to 0} (f(x) + g(x))$ ?

3 (correct answer)
2.99
5
7

Explanation: First, we estimate the individual limits from the tables. For

f(x)

, as

x \to 0

, the values approach 5, so

\lim_{x \to 0} f(x) = 5

. For

g(x)

, as

x \to 0

, the values approach -2, so

\lim_{x \to 0} g(x) = -2

. Using the limit law for sums,

\lim_{x \to 0} (f(x) + g(x)) = \lim_{x \to 0} f(x) + \lim_{x \to 0} g(x) = 5 + (-2) = 3

Question 10

Values for $f(x)$ and $g(x)$ near $x=1$ are provided. For $f(x)$ : $f(0.9) = 11.5$ , $f(0.99) = 11.95$ , $f(1.01) = 12.05$ , $f(1.1) = 12.5$ . For $g(x)$ : $g(0.9) = 2.1$ , $g(0.99) = 2.01$ , $g(1.01) = 1.99$ , $g(1.1) = 1.9$ . Estimate the value of $\lim_{x \to 1} \frac{f(x)}{g(x)}$ .

5.945
6 (correct answer)
10
24

Explanation: We first estimate the limits of the numerator and denominator separately. The table for

f(x)

suggests that

\lim_{x \to 1} f(x) = 12

. The table for

g(x)

suggests that

\lim_{x \to 1} g(x) = 2

. Since the limit of the denominator is not zero, we can use the limit law for quotients:

\lim_{x \to 1} \frac{f(x)}{g(x)} = \frac{\lim_{x \to 1} f(x)}{\lim_{x \to 1} g(x)} = \frac{12}{2} = 6

Question 11

Consider two functions, $f(x)$ and $g(x)$ . Values near $x=3$ are: $f(2.9)=5.8$ , $f(2.99)=5.98$ , $f(3.01)=6.02$ , $f(3.1)=6.2$ . And for $g(x)$ : $g(2.9)=-0.1$ , $g(2.99)=-0.01$ , $g(3.01)=0.01$ , $g(3.1)=0.1$ . What is the best description of $\lim_{x \to 3} \frac{f(x)}{g(x)}$ ?

0
6
$\infty$
Does not exist. (correct answer)

Explanation: From the tables, we estimate

\lim_{x \to 3} f(x) = 6

and

\lim_{x \to 3} g(x) = 0

. This suggests a vertical asymptote. We must check the one-sided limits of the quotient. As

x \to 3^-

f(x) \to 6

and

g(x) \to 0

through negative values. Thus,

\lim_{x \to 3^-} \frac{f(x)}{g(x)} = -\infty

. As

x \to 3^+

f(x) \to 6

and

g(x) \to 0

through positive values. Thus,

\lim_{x \to 3^+} \frac{f(x)}{g(x)} = +\infty

. Since the left and right-hand limits are not equal, the two-sided limit does not exist.

Question 12

The following table gives values for a differentiable function $f(x)$ . $f(1.99) = 7.960$ , $f(1.999) = 7.996$ , $f(2) = 8$ , $f(2.001) = 8.004$ , $f(2.01) = 8.040$ . Use this data to provide the best estimate of $f'(2)$ .

2
4 (correct answer)
8
0.004

Explanation: The derivative

f'(2)

is defined as

\lim_{h \to 0} \frac{f(2+h)-f(2)}{h}

. We can estimate this limit by calculating the slope of secant lines for small values of

h

. Using

h=0.001

, the slope is

\frac{f(2.001)-f(2)}{0.001} = \frac{8.004-8}{0.001} = \frac{0.004}{0.001} = 4

. Using

h=-0.001

, the slope is

\frac{f(1.999)-f(2)}{-0.001} = \frac{7.996-8}{-0.001} = \frac{-0.004}{-0.001} = 4

. Since both sides approach 4, this is the best estimate for

f'(2)

Question 13

A scientist records the following data for a function $y(t)$ near $t=5$ : $y(4.9)=1.9$ , $y(4.99)=1.99$ , $y(4.999)=1.999$ , $y(5.001)=4.0$ , $y(5.01)=2.01$ , $y(5.1)=2.1$ . If it is suspected that one of these measurements is erroneous, what is the most likely value of $\lim_{t \to 5} y(t)$ ?

2 (correct answer)
3
4.0
Does not exist.

Explanation: Examining the trend of the data is key. As

t

approaches 5 from the left,

y(t)

approaches 2. As

t

approaches 5 from the right, the values (2.1, 2.01) also suggest a trend toward 2. The value

y(5.001)=4.0

is a significant outlier that breaks this pattern. Assuming this point is an error, the consistent trend from both sides indicates that the limit is 2.

Question 14

Consider a function $f(x)$ with the following values: $f(-2.1) = -1.21$ , $f(-2.01) = -1.0201$ , $f(-1.99) = -0.9801$ , $f(-1.9) = -0.61$ . What is the best estimate for $\lim_{x \to -2} \frac{f(x)+1}{x+2}$ ?

-2
-1
0
2 (correct answer)

Explanation: From the table, we first estimate that

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

. The expression

\lim_{x \to -2} \frac{f(x)+1}{x+2}

is in the indeterminate form

0/0

. We can estimate the limit by calculating the value of the quotient for

x

values close to -2. For

x = -2.01

, the quotient is

\frac{f(-2.01)+1}{-2.01+2} = \frac{-1.0201+1}{-0.01} = \frac{-0.0201}{-0.01} = 2.01

. For

x = -1.99

, the quotient is

\frac{f(-1.99)+1}{-1.99+2} = \frac{-0.9801+1}{0.01} = \frac{0.0199}{0.01} = 1.99

. The values of the quotient are approaching 2 as

x

approaches -2.

Question 15

A function $f(x)$ is evaluated for several large values of $x$ : $f(10) = 3.1$ , $f(100) = 3.01$ , $f(1000) = 3.001$ , and $f(10000) = 3.0001$ . Based on this trend, what is the best estimate for $\lim_{x \to \infty} f(x)$ ?

$\infty$
3.0001
3 (correct answer)
0

Explanation: The table shows that as

x

increases without bound (10, 100, 1000, 10000), the value of

f(x)

gets progressively closer to 3. This indicates that the function has a horizontal asymptote at

y=3

. Therefore, the limit of

f(x)

x

approaches infinity is 3.

Question 16

A function $f(x)$ is evaluated at several points near $x = 2$ . The results are: $f(1.9) = 6.859$ , $f(1.99) = 6.985$ , $f(1.999) = 6.998$ , $f(2) = 10$ , $f(2.001) = 7.001$ , $f(2.01) = 7.015$ , and $f(2.1) = 7.151$ . Based on this data, what is the best estimate for $\lim_{x \to 2} f(x)$ ?

6.998
7 (correct answer)
10
The limit does not exist.

Explanation: To estimate

\lim_{x \to 2} f(x)

, we examine the values of

f(x)

x

gets arbitrarily close to 2 from both the left and the right. As

x

approaches 2 from the left (1.9, 1.99, 1.999),

f(x)

approaches 7. As

x

approaches 2 from the right (2.1, 2.01, 2.001),

f(x)

also approaches 7. Since the left-hand and right-hand limits both appear to be 7, the best estimate for the limit is 7. The value of the function at

x=2

, which is

f(2)=10

, does not affect the value of the limit.

Question 17

Let $g(x)$ be a function with values: $g(2.9)=-1.1$ , $g(2.99)=-1.01$ , $g(3.01)=-0.99$ , $g(3.1)=-0.9$ . Let $f(u)$ be a function with values: $f(-1.1)=8.7$ , $f(-1.01)=8.97$ , $f(-0.99)=9.03$ , $f(-0.9)=9.3$ . Based on this data, estimate $\lim_{x \to 3} f(g(x))$ .

-1
3
9 (correct answer)
8.97

Explanation: To evaluate the limit of the composite function

\lim_{x \to 3} f(g(x))

, we first find the limit of the inner function,

u = g(x)

, as

x \to 3

. The table for

g(x)

shows that as

x

approaches 3,

g(x)

approaches -1. So,

\lim_{x \to 3} g(x) = -1

. Now we evaluate the limit of the outer function as its input approaches -1:

\lim_{u \to -1} f(u)

. The table for

f(u)

shows that as

u

approaches -1,

f(u)

approaches 9. Therefore,

\lim_{x \to 3} f(g(x)) = 9

Question 18

A function $f(x)$ has values near $x=1$ as follows: $f(0.9) = -4.8$ , $f(0.99) = -4.98$ , $f(1.01) = -5.02$ , $f(1.1) = -5.2$ . If $k$ is a constant such that $\lim_{x \to 1} (k \cdot f(x)) = 10$ , what is the value of $k$ ?

-5
2
-2 (correct answer)
10

Explanation: First, we estimate

\lim_{x \to 1} f(x)

from the table. As

x

approaches 1 from both sides,

f(x)

approaches -5. Using the constant multiple rule for limits,

\lim_{x \to 1} (k \cdot f(x)) = k \cdot \lim_{x \to 1} f(x)

. We are given that this limit is 10. So, we have the equation

k \cdot (-5) = 10

. Solving for

k

gives

k = 10 / (-5) = -2

Question 19

The table below provides values for a rational function $h(x) = \frac{p(x)}{q(x)}$ , where $p(x)$ and $q(x)$ are polynomials. Values are: $h(1.9) = 2.45$ , $h(1.99) = 2.495$ , $h(2.01) = 2.505$ , $h(2.1) = 2.55$ . What is the most likely value of $\lim_{x \to 2} h(x)$ ?

2.495
2.5 (correct answer)
2.505
3

Explanation: The table shows that as

x

approaches 2 from the left,

h(x)

approaches 2.5. As

x

approaches 2 from the right,

h(x)

also approaches 2.5. Since the function is rational and the limit exists, the value of the limit is the value the function is approaching from both sides, which is 2.5 or 5/2.

Question 20

A table of values for a function $f(x)$ is provided: $f(-1)=5$ , $f(0)=6$ , $f(0.9)=6.9$ , $f(1.001)=7.003$ , $f(1.1)=7.3$ , $f(2)=11$ . Use the most relevant data to estimate $\lim_{x \to 1} f(x)$ .

6.9515
7 (correct answer)
7.003
Cannot be determined.

Explanation: To estimate the limit as

x \to 1

, we must consider the values of

f(x)

for

x

closest to 1. The most relevant data points are for

x=0.9

(just to the left of 1) and

x=1.001

(just to the right of 1). The corresponding function values are

f(0.9)=6.9

and

f(1.001)=7.003

. These values suggest that as

x

gets closer to 1,

f(x)

gets closer to 7. The other data points are too far from

x=1

to be as relevant.