Gateway Exam period
extended until Dec. 3
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From the chain rule:
d
sin (3t2 ) = cos (3t2 )(6t)
dx
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From the chain rule:
d
sin (3t2 ) = cos (3t2 )(6t)
dx
When we integrate
Z
cos (3t2 )(6t) dt
we observe that cos (3t2 )(6t) has the basic form cos (u)u0 .
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From the chain rule:
d
sin (3t2 ) = cos (3t2 )(6t)
dx
When we integrate
Z
cos (3t2 )(6t) dt
we observe that cos (3t2 )(6t) has the basic form cos (u)u0 .
This insight is facilitated by the substitution technique.
We set u = 3t2 and compute du = 6t dt. Now making the change
of variable from x to u, we get
Z
2
cos (3t )(6t) dt =
Z
sin u du = cos u + C = cos (3t2 ) + C
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R
Integrate 3x21+5 dx
First try: Set u = 3x2 + 5 and du = 6x dx. Then
Z
Z
1 1
1
dx =
du
2
3x + 5
u 6x
We cannot eliminate all the x0 s.
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3 / 26
R
Integrate 3x21+5 dx
First try: Set u = 3x2 + 5 and du = 6x dx. Then
Z
Z
1 1
1
dx =
du
2
3x + 5
u 6x
We cannot eliminate all the x0 s.
Second try:
Z
1
dx =
3x2 + 5
Z
1
1
1
( )3 2
dx =
5 5x + 1
5
Z
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1
dx
+1
3 2
x
5
3 / 26
R
Integrate 3x21+5 dx
First try: Set u = 3x2 + 5 and du = 6x dx. Then
Z
Z
1 1
1
dx =
du
2
3x + 5
u 6x
We cannot eliminate all the x0 s.
Second try:
Z
1
dx =
3x2 + 5
√
3
Set u = √ x and du =
5
Z
1
1
dx =
3 2
5
x +1
5
Z
Z
1
1
1
1
( )3 2
dx =
dx
3 2
5 5x + 1
5
x
+
1
5
√
√
3
5
√ dx =⇒ √ du = dx
5
3
√
√ Z
Z
1
1
1
5
5
√ du = √
du
2
2
5
u +1 3
u +1
5 3
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To finish
√ Z
√
5
5
1
√
du = √ tan−1 (u) + C
2
u +1
5 3
5 3
√
3
Substituting u = √ x to return to the original variable x:
5
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To finish
√ Z
√
5
5
1
√
du = √ tan−1 (u) + C
2
u +1
5 3
5 3
√
3
Substituting u = √ x to return to the original variable x:
5
√
√
5
3
= √ tan−1 ( √ x) + C
5 3
5
√
√
Z
1
5
3
−1
dx = √ tan ( √ x) + C
2
3x + 5
5 3
5
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4 / 26
Integrate
R
2x+1
3x2 +5
dx
First try: Set u = 3x2 + 5 and du = 6x dx. Then
Z
Z
2x + 1
1
dx =
(??) du
2
3x + 5
u
The +1 in the numerator prevent this substitution from working.
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5 / 26
Integrate
R
2x+1
3x2 +5
dx
First try: Set u = 3x2 + 5 and du = 6x dx. Then
Z
Z
2x + 1
1
dx =
(??) du
2
3x + 5
u
The +1 in the numerator prevent this substitution from working.
Second try: We rewrite the integral as
Z
Z
Z
2x + 1
2x
1
dx =
dx +
dx
2
2
2
3x + 5
3x + 5
3x + 5
Z
2x
2
The substitution u = 3x + 5 works for
dx
3x2 + 5
√
Z
1
3
dx is just the previous example using u = √ x
2
3x + 5
5
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Clicker Question
Use substitution to find
(a) ln | sec x| + C
(d) All of the above
R
tan x dx. (Hint: Write tan x =
sin x
)
cos x
1
|+C
(c) − ln | cos x| + C
cos x
(e) None of these
(b) ln |
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6 / 26
Clicker Question
Use substitution to find
(a) ln | sec x| + C
(d) All of the above
R
tan x dx. (Hint: Write tan x =
sin x
)
cos x
1
|+C
(c) − ln | cos x| + C
cos x
(e) None of these
(b) ln |
Let u = cos x. Then du = − sin x dx =⇒ −du = sin x dx.
Z
Z
Z
sin x
1
dx =
(−du) = − u−1 du = − ln |u| + C
cos x
u
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Clicker Question
Use substitution to find
(a) ln | sec x| + C
(d) All of the above
R
tan x dx. (Hint: Write tan x =
sin x
)
cos x
1
|+C
(c) − ln | cos x| + C
cos x
(e) None of these
(b) ln |
Let u = cos x. Then du = − sin x dx =⇒ −du = sin x dx.
Z
Z
Z
sin x
1
dx =
(−du) = − u−1 du = − ln |u| + C
cos x
u
Returning to our original variable u = u = cos x, we see that
Z
tan x dx = − ln | cos x| + C
OR
Z
tan x dx = ln |
1
| + C = ln | sec x| + C
cos x
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6 / 26
Substitution Rule for Definite Integrals
Using the substitution u = g(x) we get
Z
b
0
Z
g(b)
f (g(x))g (x) dx =
a
f (u) du
g(a)
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Exam period extended until Dec. 3
7 / 26
Substitution Rule for Definite Integrals
Using the substitution u = g(x) we get
b
Z
Z
0
g(b)
f (g(x))g (x) dx =
a
f (u) du
g(a)
Z
Example:
1
1
18x2 (6x3 + 5) 4 dx
0
du
= 18x2 =⇒ du = 18x2 dx
dx
Z 1
Z g(1)
Z 11
1
1
1
2
3
18x (6x + 5) 4 dx =
(u) 4 du =
(u) 4 du
We take u = g(x) = 6x3 + 5 and
0
g(0)
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5
7 / 26
Substitution Rule for Definite Integrals
Using the substitution u = g(x) we get
b
Z
Z
0
g(b)
f (g(x))g (x) dx =
a
f (u) du
g(a)
Z
Example:
1
1
18x2 (6x3 + 5) 4 dx
0
du
= 18x2 =⇒ du = 18x2 dx
dx
Z 1
Z g(1)
Z 11
1
1
1
2
3
18x (6x + 5) 4 dx =
(u) 4 du =
(u) 4 du
We take u = g(x) = 6x3 + 5 and
0
Z
g(0)
1
2
3
1
4
Z
18x (6x + 5) dx =
0
5
11
5
1
4
4
4
5/4
(u) 4 du = u5/4 |11
− (5)5/4
5 = (11)
5
5
5
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7 / 26
Why the Substitution Rule for Definite Integrals works
If F 0 = f then
Z
d
F (g(x)) = f (g(x))g 0 (x) and hence
dx
f (g(x))g 0 (x) dx = F (g(x)) + C and
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8 / 26
Why the Substitution Rule for Definite Integrals works
d
F (g(x)) = f (g(x))g 0 (x) and hence
dx
If F 0 = f then
Z
f (g(x))g 0 (x) dx = F (g(x)) + C and
Z
b
f (g(x))g 0 (x) dx = F (g(b)) − F (g(a))
a
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Why the Substitution Rule for Definite Integrals works
d
F (g(x)) = f (g(x))g 0 (x) and hence
dx
If F 0 = f then
Z
f (g(x))g 0 (x) dx = F (g(x)) + C and
Z
b
f (g(x))g 0 (x) dx = F (g(b)) − F (g(a))
a
Which is the same after substitution as
Z
Z
g(b)
f (u) du = F (g(b)) − F (g(a))
f (u) du = F (u) + C and
g(a)
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8 / 26
Clicker Question
Z
Evaluate
π/2
(1 − sin2 (x)) cos (x) dx
0
(a) 0
(b) 1/3
(c) 2/3
(d)
1
(e) 3/8
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9 / 26
Clicker Question
Z
Evaluate
π/2
(1 − sin2 (x)) cos (x) dx
0
(a) 0
(b) 1/3
(c) 2/3
(d)
1
(e) 3/8
du
= cos x =⇒ du = cos x dx.
dx
When x = 0, u = sin (0) = 0 and when x = π/2, u = sin (π/2) = 1.
Let u = sin x. Then
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9 / 26
Clicker Question
Z
Evaluate
π/2
(1 − sin2 (x)) cos (x) dx
0
(a) 0
(b) 1/3
(c) 2/3
(d)
1
(e) 3/8
du
= cos x =⇒ du = cos x dx.
dx
When x = 0, u = sin (0) = 0 and when x = π/2, u = sin (π/2) = 1.
Z π/2
Z 1
2
(1 − sin (x)) cos (x) dx =
(1 − u2 ) du
Let u = sin x. Then
0
0
1
1
2
= (u − u3 )|10 = (1 − ) − 0 =
3
3
3
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9 / 26
Interpretations of the Definite Integral
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10 / 26
Given a function f (x) that is continuous on the interval [a,b] we
divide the interval into n subintervals of equal width, ∆x, and
from each interval choose a point x̄i .
Definition
The definite integral of f (x) from a to b is
Z
b
f (x) dx = lim
a
n→∞
n
X
f (x̄i )∆x
i=1
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11 / 26
Given a function f (x) that is continuous on the interval [a,b] we
divide the interval into n subintervals of equal width, ∆x, and
from each interval choose a point x̄i .
Definition
The definite integral of f (x) from a to b is
Z
b
f (x) dx = lim
a
n→∞
n
X
f (x̄i )∆x
i=1
We can think of the definite integral as
equal to the net area
Area above the x-axis
minus
Area below the x-axis
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11 / 26
Let r(t) be the rate at which the worlds oil is consumed, where t
is measured in years starting at t = 0 on Jan 1, 2000. r(t) is
measured in barrels per year.
Z
What does
8
r(t) dt represent?
0
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12 / 26
Let r(t) be the rate at which the worlds oil is consumed, where t
is measured in years starting at t = 0 on Jan 1, 2000. r(t) is
measured in barrels per year.
Z
What does
8
r(t) dt represent?
0
Let V (t) represent the volume of oil consumed. Then V 0 (t) = r(t).
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12 / 26
Let r(t) be the rate at which the worlds oil is consumed, where t
is measured in years starting at t = 0 on Jan 1, 2000. r(t) is
measured in barrels per year.
Z
What does
8
r(t) dt represent?
0
Let V (t) represent the volume of oil consumed. Then V 0 (t) = r(t).
By the Fundamental Theorem,
Z
V (8) − V (0) =
8
r(t) dt
0
which is the number of barrels of oil consumed between Jan 1,
2000 and Dec. 31, 2008.
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Z
Another interpretation of
b
F 0 (x) dx:
a
F (b) − F (a) equals the net change in F (t) between t = a and
t = b.
Thus the net change in F over [a, b] is equal to
Z b
F (b) − F (a) =
F 0 (x) dx
a
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13 / 26
Z
Another interpretation of
b
F 0 (x) dx:
a
F (b) − F (a) equals the net change in F (t) between t = a and
t = b.
Thus the net change in F over [a, b] is equal to
Z b
F (b) − F (a) =
F 0 (x) dx
a
When we compute the definite integral of a rate of change and we
get the net change in the quantity.
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Z
Another interpretation of
b
F 0 (x) dx:
a
F (b) − F (a) equals the net change in F (t) between t = a and
t = b.
Thus the net change in F over [a, b] is equal to
Z b
F (b) − F (a) =
F 0 (x) dx
a
When we compute the definite integral of a rate of change and we
get the net change in the quantity.
Example: If s(t) is the distance in miles a car is from home and
v(t) = s0 (t) is the velocity (t in hours),
the distance traveled in
Z
2
the first 2 hours is s(2) − s(0) =
v(t) dt.
0
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13 / 26
Another Example:
F (t) represents a bacteria population which is 5 million at t = 0
hours. It is growing at a rate of 2t million per hour. Find the
increase in the population during the first hour.
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Exam period extended until Dec. 3
14 / 26
Another Example:
F (t) represents a bacteria population which is 5 million at t = 0
hours. It is growing at a rate of 2t million per hour. Find the
increase in the population during the first hour.
Z
The change in population is F (1) − F (0) =
1
F 0 (t) dt.
0
We are given that the rate of change of F is F 0 (t) = 2t .
Z 1
Hence the change in population is F (1) − F (0) =
2t dt.
0
To compute the population after 1 hour, we need to compute
F (1).
Z
F (1) = F (0) +
1
2t dt ≈ 5 + 1.44 = 6.44 million
0
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14 / 26
Clicker Question
Suppose water is being drained out of a tank at a rate of t2 + 1
gal/min. How much water is drained out in the first 3 minutes?
(a) 10 gal
(b) 12 gal
(c) 15 gal (d)
30 gal
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(e) 20 gal
15 / 26
Clicker Question
Suppose water is being drained out of a tank at a rate of t2 + 1
gal/min. How much water is drained out in the first 3 minutes?
(a) 10 gal
(b) 12 gal
(c) 15 gal (d)
30 gal
(e) 20 gal
If V (t) is the volume of water in a tank then V 0 (t) = t2 + 1
gal/min is the rate
Z of change in V (t) and
3
t2 + 1 dt is the net change in the volume as we
V (3) − V (0) =
0
go from time t = 0 to time t = 3 . The volume of water drained
out in the first 3 minutes is
Z
V (3) − V (0) =
0
3
t2 + 1 dt = (
t3
+ t)|30 = 27/3 + 3 − 0 = 12 gal
3
Answer = b
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15 / 26
Distance traveled
If an object travels at a constant velocity of 5 ft/sec for 20
seconds, then it will have traveled (5)(20)=100 feet. When
velocity is constant,
distance = velocity × time elapsed
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Exam period extended until Dec. 3
16 / 26
Distance traveled
If an object travels at a constant velocity of 5 ft/sec for 20
seconds, then it will have traveled (5)(20)=100 feet. When
velocity is constant,
distance = velocity × time elapsed
v
When velocity is v = 6 the distance
traveled is (6)(2), the area of the
rectangular area under the velocity
graph.
t
Distance traveled = net area between graph of velocity and t-axis
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16 / 26
If s(t) is the function giving the position of some object at time t
we know that the velocity of the object at any time t is
v(t) = s0 (t).
The displacement of the object time
t = 0 to time t = 5 is,
Z 5
s(5) − s(0) =
v(t) dt
0
v(t)
t
1
4 5
Which equals
Distance traveled in positive direction (over [0,1] and [4,5])
minus
Distance traveled in negative direction (over [1,4]).
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If v(t) is both positive and negative (i.e. the object moves to both
the right and left) in the time frame, this will NOT give the
total distance traveled. It will only give the displacement,
i.e. the difference between where the object started and where it
ended up.
Total distance traveled equals
Distance traveled in positive direction
plus
Distance traveled in negative direction
|v(t)|
t
1
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4 5
18 / 26
If v(t) is both positive and negative (i.e. the object moves to both
the right and left) in the time frame, this will NOT give the
total distance traveled. It will only give the displacement,
i.e. the difference between where the object started and where it
ended up.
Total distance traveled equals
|v(t)|
Distance traveled in positive direction
plus
t
Distance traveled in negative direction
1
4 5
Z 5
Z 1
Z 4
Z 5
|v(t)| dt =
|v(t)| dt +
|v(t)| dt +
|v(t)| dt
0
Z
0
1
4
5
|v(t)| dt = (s(1) − s(0)) + |s(4) − s(1)| + (s(5) − s(4))
0
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Clicker Question
Z
3
|3t − 5| dt.
Evaluate
0
(a) 0
(b) 41/6
(c) 37/3 (d)
59/2
(e) None of these
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19 / 26
Clicker Question
Z
3
|3t − 5| dt.
Evaluate
0
(a) 0
(b) 41/6
(c) 37/3 (d)
59/2
(e) None of these
We need to determine where the quantity on the inside of the
absolute value bars is negative and where it is positive
5
3(t − ) > 0 if t >
3
5
3(t − ) < 0 if t <
3
5
3
5
3
=⇒ |3t − 5| =

 3t − 5
if t ≥
5
3
5 − 3t
if t <
5
3

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19 / 26
Clicker Question
Z
3
|3t − 5| dt.
Evaluate
0
(a) 0
(b) 41/6
(c) 37/3 (d)
59/2
(e) None of these
We need to determine where the quantity on the inside of the
absolute value bars is negative and where it is positive
5
3(t − ) > 0 if t >
3
5
3(t − ) < 0 if t <
3
Z 3
Z
|3t − 5| dt =
0
5
3
5
3
0
=⇒ |3t − 5| =

 3t − 5
if t ≥
5
3
5 − 3t
if t <
5
3

5/3
Z
3
5 − 3t dt +
3t − 5 dt
5/3
3
3
5/3
= (5t − t2 )|0 + (( t2 − 5t)|35/3
2
2
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Answer = b
19 / 26
Review of L0 Hopital”s Rule
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20 / 26
Clicker Question
ex − 1
.
x→0
x
Evaluate lim
(a) 0
(b) 1
(c) −1
(d)
e
(e) None of these
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Clicker Question
ex − 1
.
x→0
x
Evaluate lim
(a) 0
(b) 1
(c) −1
(d)
e
(e) None of these
ex − 1
limx→0 ex − 1
0
=
=
x→0
x
limx→0 x
0
lim
Apply L0 Hopital0 s Rule:
ex − 1
ex
= lim
=1
x→0
x→0 1
x
lim
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21 / 26
Clicker Question
Evaluate the appropriate limit to find the right horizontal
asymptote of y = (3x)1/x .
(a) y = 0
(b) y = 1
None of these
(c) y = 1/3
(d)
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y=3
(e)
22 / 26
Clicker Question
Evaluate the appropriate limit to find the right horizontal
asymptote of y = (3x)1/x .
(a) y = 0
(b) y = 1
None of these
(c) y = 1/3
(d)
y=3
(e)
Take ln y = ln (3x)1/x and find
( lim ln y)
lim ln y which we then use to find lim y = lim eln y = e x→∞
x→∞
x→∞
x→∞
lim ln (3x)
∞
ln (3x)
= x→∞
=
x→∞
x
lim x
∞
lim ln y = lim ln (3x)1/x = lim
x→∞
x→∞
x→∞
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Use L0 Hopital0 s Rule:
1/x
ln (3x)
= lim
=0
x→∞ 1
x→∞
x
lim
( lim ln y)
lim y = lim eln y = e x→∞
= e0 = 1
x→∞
x→∞
y = 1 is the right horizontal asymptote
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Evaluate the appropriate
√ limit to find the right horizontal
asymptote of y = x − x2 − 3x.
lim x −
x→∞
√
x2 − 3x = ∞ − ∞ an indeterminate form.
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24 / 26
Evaluate the appropriate
√ limit to find the right horizontal
asymptote of y = x − x2 − 3x.
lim x −
√
x→∞
x−
√
x2
− 3x = (x −
lim (x −
x→∞
x2 − 3x = ∞ − ∞ an indeterminate form.
√
√
x2
√
x2 − (x2 − 3x)
x + x2 − 3x
√
√
=
− 3x)
x + x2 − 3x
x + x2 − 3x
∞
3x
√
=
2
x→∞ x +
∞
x − 3x
x2 − 3x) = lim
We can now apply L0 Hopital0 s Rule
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24 / 26
Clicker Question
Evaluate lim (x −
x→∞
(a) 0
(b) 1
√
x2 − 3x) = lim
(c) 2/3
3x
√
x→∞ x +
x2 − 3x
(d)
3/2
(e) None of these
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()
Exam period extended until Dec. 3
25 / 26
Clicker Question
Evaluate lim (x −
x→∞
(a) 0
(b) 1
√
x2 − 3x) = lim
(c) 2/3
3x
√
x→∞ x +
x2 − 3x
(d)
3/2
(e) None of these
We could apply L0 Hopital0 s Rule:
√
3x
√
lim (x − x2 − 3x) = lim
x→∞
x→∞ x +
x2 − 3x
3x
3
√
= lim
lim
2
x→∞ x +
x − 3x x→∞ 1 + 2√2x+3
x2 −3x
But use some algebra
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25 / 26
3x
√
x→∞ x +
x2 − 3x
lim
√
3x
1/x
3x
√
√
=(
)
2
1/x x + x2 − 3x
x + x − 3x
3
=
√
1+
2
x
√ −3x
x2
=
x > 0 =⇒
x2 = x
3
1+
p
1 − 3/x
Now we can finish the limit
lim (x −
x→∞
√
3x
√
x→∞ x +
x2 − 3x
x2 − 3x) = lim
3
3
p
=
x→∞ 1 +
2
1 − 3/x
= lim
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()
Exam period extended until Dec. 3
26 / 26