This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for
the skipped part, and you will not be able to come back to the skipped part.
If an arrow is shot upward on Mars with a speed of 61 m/s, its height in meters t seconds later is given by
Exercise (a)
Find the average speed over the given time intervals.
(i) [1, 2]
(ii) [1, 1.5]
(iii) [1, 1.1]
(iv) [1, 1.01]
(v) [1, 1.001]
Exercise (b)
Estimate the speed when t = 1.
Step 1
To estimate the speed when t = 1, we need to estimate when h approaches . This yields
The displacement (in centimeters) of a particle moving back and forth along a straight line is given by the equation of motion
where t is measured in seconds. (Round your answers to two decimal places.)
(a) Find the average velocity during each time period.
(i) [1, 2]
cm/s
(ii) [1, 1.1]
cm/s
(iii) [1, 1.01]
cm/s
(iv) [1, 1.001]
cm/s
(b) Estimate the instantaneous velocity of the particle when t = 1.
cm/s
y = 61t − 1.86t2.
vave = 57.28 − 1.86h
vave = m/s.
s = 3 sin(πt) + 4 cos(πt),
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Video Example
EXAMPLE 1 Find an equation of the tangent line to the function at the
point P(1, 3).
SOLUTION We will be able to find an equation of the tangent line t as soon as
we know its slope m. The difficulty is that we know only one point, P, on t,
whereas we need two points to compute the slope. But observe that we can
compute an approximation to m by choosing a nearby point on the
graph (as in the figure) and computing the slope mPQ of the secant line PQ.
[A secant line, from the Latin word secans, meaning cutting, is a line that cuts
(intersects) a curve more than once.]
We choose so that Then,
For instance, for the point Q(1.5, 6.75) we have
The tables below show the values of mPQ for several values of x close to 1. The
closer Q is to P, the closer x is to 1 and, it appears from the tables, the
closer mPQ is to . This suggests that the slope of the tangent
line t should be m = .
x mPQ x mPQ
2 9 0 3
1.5 7.5 .5 4.5
1.1 6.3 .9 5.700
1.01 6.030 .99 5.970
1.001 6.003 .999 5.997
We say that the slope of the tangent line is the limit of the slopes of the secant
lines, and we express this symbolically by writing
Assuming that this is indeed the slope of the tangent line, we use the point-slope
form of the equation of a line (see Appendix B) to write the equation of the
tangent line through (1, 3) as
or
The graphs below illustrate the limiting process that occurs in this example.
As Qapproaches P along the graph, the corresponding secant lines rotate
about P and approach the tangent line t.
y = 3x
2
Q(x, 3x
2)
x ≠ 1 Q ≠ P.
mPQ = . 3x
2 − 3
x − 1
mPQ = = = .
− 3
− 1 .5
lim mPQ = m and = .
Q → P
lim
x → 1
3x
2 − 3
x − 1
y − = (x − 1)
y = x − .
This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for
the skipped part, and you will not be able to come back to the skipped part.
Tutorial Exercise
Find an equation of the line passing through the given points.
Explain what it means to say that
In this situation is it possible that exists? Explain.
For the function h whose graph is given, state the value of each quantity, if it exists. (If an answer does not exist, enter DNE.)
(a)
(b)
(1, 1), 6, −
2
3
lim f(x) = 1 and f(x) = 2.
x → 6−
lim
x → 6+
As x approaches 6, f(x) approaches 1, but f(6) = 2.
As x approaches 6 from the right, f(x) approaches 1. As x approaches 6 from the left, f(x) approaches 2.
As x approaches 6 from the left, f(x) approaches 1. As x approaches 6 from the right, f(x) approaches 2.
As x approaches 6, f(x) approaches 2, but f(6) = 1.
lim f(x)
x → 6
Yes, f(x) could have a hole at (6, 1) and be defined such that f(6) = 2.
Yes, f(x) could have a hole at (6, 2) and be defined such that f(6) = 1.
Yes, if f(x) has a vertical asymptote at x = 6, it can be defined sulcimhf(txh)at f(=x)1, = 2f,(axn) d exists.
x →6−
lim
x →6+
lim
x →6
limf(xN)o, cannot exf(isxtf)(ixf) ≠ .
x →6
lim
x →6−
lim
x →6+
lim h(x)
x → −3−
lim h(x)
x → −3+ - –/4 POINTS SCALCET8 2.1.JIT.003.MI.SA. MY NOTES ASK YOUR TEACHER
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(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
For the function f whose graph is shown, state the following. (If an answer does not exist, enter DNE.)
(a)
(b)
lim h(x)
x → −3
h(−3)
lim h(x)
x → 0−
lim h(x)
x → 0+
lim h(x)
x → 0
h(0)
lim h(x)
x → 2
h(2)
lim h(x)
x → 5+
lim h(x)
x → 5−
lim f(x)
x → −7
lim f(x)
x → −3 - –/3 POINTS SCALCET8 2.2.009. MY NOTES ASK YOUR TEACHER
(c)
(d)
(e)
(f) The equations of the vertical asymptotes.
x = (smallest value)
x =
x =
x = (largest value)
This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for
the skipped part, and you will not be able to come back to the skipped part.
Tutorial Exercise
A patient receives a 150 mg injection of a drug every 4 hours. The graph shows the amount f(t) of the drug in the bloodstream
after thours.
Find and
lim f(x)
x → 0
lim f(x)
x → 6−
lim f(x)
x → 6+
lim f(t)
t → 8−
lim f(t).
t → 8+ - –/4 POINTS SCALCET8 2.2.010.MI.SA. MY NOTES ASK YOUR TEACHER
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Sketch the graph of the function.
Use the graph to determine the values of a for which does not exist. (Enter your answers as a comma-separated list.)
A graphing calculator is recommended.
Use the graph of the function f to state the value of each limit, if it exists. (If an answer does not exist, enter DNE.)
f(x) =
4 + x if x < −2
x
2 if −2 ≤ x < 2
6 − x if x ≥ 2
lim f(x)
x → a
a =
f(x) =
7
1 + e
1/x - –/3 POINTS SCALCET8 2.2.013. MY NOTES ASK YOUR TEACHER
(a)
(b)
(c)
Evaluate the function f(x) at the given numbers (correct to six decimal places).
x f(x)
4.1
4.05
4.01
4.001
4.0001
x f(x)
3.9
3.95
3.99
3.999
3.9999
Guess the value of the limit (correct to six decimal places). (If an answer does not exist, enter DNE.)
Use a table of values to estimate the value of the limit. If you have a graphing device, use it to confirm your result graphically. (Round
your answer to two decimal places.)
lim f(x)
x →0−
lim f(x)
x →0+
lim f(x)
x →0
f(x) = ,
x
2 − 4x
x
2 − 16
x = 4.1, 4.05, 4.01, 4.001, 4.0001,
3.9, 3.95, 3.99, 3.999, 3.9999
lim
x →4
x
2 − 4x
x
2 − 16
x
2 lim ln(x)
x →0+ - –/4 POINTS SCALCET8 2.2.019. MY NOTES ASK YOUR TEACHER
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Determine the infinite limit.
∞
−∞
Determine the infinite limit.
∞
−∞
Evaluate the function for values of x that approach 1 from the left and from the right.
Given that
find the limits, if they exist. (If an answer does not exist, enter DNE.)
(a)
(b)
(c)
ln(x lim 2 − 9)
x →3+
lim
x →2+
x
2 − 2x
x
2 − 4x + 4
f(x) =
8
x
3 − 1
lim f(x) =
x →1−
lim f(x) =
x →1+
lim f(x) = 9 g(x) = −2 h(x) = 0,
x →3
lim
x →3
lim
x →3
lim [f(x) + 4g(x)]
x →3
[g(x)] lim 3
x →3
lim
x →3
f(x) - –/3 POINTS SCALCET8 2.2.035. MY NOTES ASK YOUR TEACHER
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(d)
(e)
(f)
Evaluate the limit, if it exists. (If an answer does not exist, enter DNE.)
This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for
the skipped part, and you will not be able to come back to the skipped part.
Tutorial Exercise
Evaluate the limit, if it exists. (If an answer does not exist, enter DNE.)
Evaluate the limit, if it exists. (If an answer does not exist, enter DNE.).
Evaluate the limit, if it exists. (If an answer does not exist, enter DNE.)
lim
x →3
4f(x)
g(x)
lim
x →3
g(x)
h(x)
lim
x →3
g(x)h(x)
f(x)
lim
t →−3
t2 − 9
2t2 + 7t + 3
lim
h →0
(2 + h)3 − 8
h
lim
h → 0
− 2
h
4 + h
lim
x →−12
− 13
x + 12
x
2 + 25 - –/3 POINTS SCALCET8 2.3.015. MY NOTES ASK YOUR TEACHER
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A graphing calculator is recommended.
Use the Squeeze Theorem to show that Illustrate by graphing the functions
on the same screen.
Let and Then ⇒
Since by the Squeeze Theorem we have
Find the limit, if it exists. (If an answer does not exist, enter DNE.)
This question has several parts that must be completed sequentially. If you skip a part of the question, you will not receive any points for
the skipped part, and you will not be able to come back to the skipped part.
Tutorial Exercise
Find the limit, if it exists.
Let
(a) Find the following limits. (If an answer does not exist, enter DNE.)
(b) Does exist?
Yes
No
(x
2 cos(14πx)) = 0.
lim
x → 0
f(x) = −x
2, g(x) = x
2 cos(14πx), and h(x) = x
2
f(x) = −x
2, g(x) = x
2 cos(14πx), h(x) = x - ? ≤ cos(14πx) ≤ ? ? ≤ x
2 cos(14πx) ≤ ? .
lim f(x) = h(x) = ,
x →0
lim
x →0
lim g(x) = .
x →0
lim (6x + |x − 8|)
x →8
lim
x →−4
5x + 20
|x + 4|
f(x) = .
x
2 + 1 if x < 1
(x − 2)2 if x ≥ 1
f(x) =
f(x) =
lim
x → 1−
lim
x → 1+
lim f(x)
x → 1 - –/3 POINTS SCALCET8 2.3.035. MY NOTES ASK YOUR TEACHER
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(c) Sketch the graph of f.
If , evaluate the following limits.
(a)
(b)
lim = 1
x →0
f(x)
x
2
lim f(x)
x →0
lim
x →0
f(x)
x - –/3 POINTS SCALCET8 2.3.060. MY NOTES ASK YOUR TEACHER
Find the number a such that the limit exists.
a =
Find the value of the limit.
lim
x →−2
3x
2 + ax + a + 9
x
2 + x − 2 - –/3 POINTS SCALCET8 2.3.065. MY NOTES ASK YOUR TEACHER
Sample Solution
