Azərbaycan Dövlət Neft və Sənaye Universiteti
İMTAHAN BİLETİ № 13
Fakültə:
İTİF
Fənn:
Calculus I
Qrup:
604.22E
Ixtisas:
050509
Tələbə: Xəlilov Kamran
FIRST:
Find the limit:
√𝑥 + 6 − 𝑥
𝑥→3 𝑥 3 − 3𝑥 2
lim
SOLUTION:
First of all we check direct substitution:
√3+6−3
3
𝑥→3 3 −3∗32
Lim
= lim
3−3
𝑥→3 27−27
= lim
0
𝑥→3 0
By direct substitution, we obtain the indeterminate form 0⁄0
lim 𝑥 3 − 3𝑥 2 = 0
lim √𝑥 + 6 − 𝑥 = 0
𝑥→3
𝑥→3
In this case, you can rewrite the fraction by rationalizing the numerator:
𝑥 + 6 − 𝑥2
√𝑥 + 6 − 𝑥
√𝑥 + 6 − 𝑥 √𝑥 + 6 + 𝑥
=( 3
)(
)=
𝑥 3 − 3𝑥 2
𝑥 − 3𝑥 2
(𝑥 3 − 3𝑥 2 )(√𝑥 + 6 + 𝑥)
√𝑥 + 6 + 𝑥
=
−(𝑥 − 3)(𝑥 + 2)
𝑥 2 (𝑥 − 3)(√𝑥 + 6 + 𝑥)
lim
−(𝑥 + 2)
𝑥→3 𝑥 2 (√𝑥
Answer:
The limit of function is −
5
54
+ 6 + 𝑥)
=
−(𝑥 + 2)
𝑥 2 (√𝑥 + 6 + 𝑥)
−5
5
=−
𝑥→3 9(3 + 3)
54
= lim
Second:
Find the distance between the points (3,0) and (0,3) along the curve 𝑦 2 + 𝑥 2 = 9.
Solution:
First of all we based on (𝑦 − 𝑦0 )2 + (𝑥 − 𝑥0 )2 = 𝑅2 -we can say that center of
this circle is in (0,0), and its radius is 3. Let's sketch:
4
3
2
1
0
-4
-3
-2
-1
0
1
2
3
4
-1
-2
-3
-4
Length of circle equal to 2𝜋𝑟 and the distance between the points (3,0) and (0,3) is
equal to 25% of length of circle. That is why we firstly should carculate the length
of circle and then we should carculate 25% of length of circle:
𝜋 ≈ 3,14
2𝜋𝑟 = 2 ∗ 3,14 ∗ 3 = 18,84
1
1
4
4
Distance = ∗ 2𝜋𝑟 = ∗ 18,84 = 4,71
Answer:4,71
Third:
Use the Intermediate Value Theorem to show that there is a root of the equation in
the given interval:
(a) 𝑥 5 − 𝑥 3 + 3𝑥 − 5 = 0, (1,2).
(b) 2 sin 𝑥 = 3 − 2𝑥, (0,1).
Solution:
Intermediate Value Theorem:
If f(x) is continuous on the closed interval [a,b], f(a)≠f(b), and L is any number
between f(a) and f(b), then there is at least one number c in (a,b) such that f(c)=L.
The Intermediate Value Theorem guarantees the existence of at least one number c
in the closed interval [a,b], but it does not provide a method for finding c.
(a) 𝑥 5 − 𝑥 3 + 3𝑥 − 5 = 0, (1,2).
First of all this function is continuous on the open interval (1,2)
𝑓(𝑥) = 𝑥 5 − 𝑥 3 + 3𝑥 − 5
15 − 13 + 3 ∗ 1 − 5 = −2
25 − 23 + 3 ∗ 2 − 5 = 25
f(1)=-2
f(2)=25
f(1)<0 and f(2)>0 ,for that reason there is a root of the function in the given
interval.
(b) 2 sin 𝑥 = 3 − 2𝑥, (0,1).
This function is continuous on the open interval (1,2)
2 sin 𝑥 = 3 − 2𝑥
𝑝(𝑥) = 2 sin 𝑥 − 3 + 2𝑥
2 sin(1°) − 3 + 2 ∗ 1 ≈ −0,965
2 sin(2°) − 3 + 2 ∗ 2 ≈ 1,07
P(1)<0 and p(2)>0,that is why there is a root of the function in the given
interval.
Fourth:
Find the vertical asymptotes (if any) of the graph of the function
𝑓(𝑥) =
4
(𝑥 − 2)3
Solution:
Vertical Asymptote Theorem:
Let f and g be continuous on an open interval containing c.
If f(c)≠0, g(c)=0, and there exists an open interval containing c such that
g(x)≠0 for all x in the interval, then the graph of the function given by
h(x)=f(x)/g(x) has a vertical asymptote at x=c.
Let’s show this condition by limit process:
𝑙𝑖𝑚𝑓(𝑥) = ∞
𝑥→𝑎
𝑙𝑖𝑚𝑓(𝑥) = −∞
𝑥→𝑎
If f(x) goes to infinity (or negative infinity) as x approaches c from the right or
the left, then the line x=c is a vertical asymptote of the graph of f.
𝑓(𝑥) =
4
𝑔(𝑥)
=
(𝑥 − 2)3 ℎ(𝑥)
𝑔(𝑥) ≠ 0
İf c=2, h(c)=0 in this case vertical asymptote is x=2
6
4
2
0
-30
-20
-10
0
10
20
30
-2
-4
-6
Answer: The vertical asymptote of the graph is x=2
Fifth:
Prove the differentiability rule:
d
d
[𝑓(𝑥)]𝑔(𝑥) −
[𝑔(𝑥)]𝑓(𝑥)
d 𝑓(𝑥)
d𝑥
d𝑥
[
]=
d𝑥 𝑔(𝑥)
𝑔2 (𝑥)
Solution:
𝒇(𝒙)
𝐥𝐢𝐦 (
) = 𝐥𝐢𝐦
∆𝒙→𝟎 𝒈(𝒙)
∆𝒙→𝟎
𝐥𝐢𝐦
∆𝒙→𝟎
𝒇(𝒙 + ∆𝒙) 𝒇(𝒙)
−
𝒈(𝒙 + ∆𝒙) 𝒈(𝒙)
∆𝒙
=
𝒇(𝒙 + ∆𝒙) 𝒈(𝒙) 𝒇(𝒙) 𝒈(𝒙 + ∆𝒙)
𝟏
∙(
−
) =
∆𝒙 𝒈(𝒙 + ∆𝒙) 𝒈(𝒙) 𝒈(𝒙) 𝒈(𝒙 + ∆𝒙)
𝒇(𝒙 + ∆𝒙)𝒈(𝒙) − 𝒇(𝒙)𝒈(𝒙) + 𝒇(𝒙)𝒈(𝒙) − 𝒇(𝒙)𝒈(𝒙 + ∆𝒙)
𝐥𝐢𝐦 (
)=
∆𝒙𝒈(𝒙)𝒈(𝒙 + ∆𝒙)
∆𝒙→𝟎
𝑔(𝑥)(𝑓(𝒙 + ∆𝒙) − 𝒇(𝒙)) − 𝑓(𝑥)(𝑔(𝑥) − 𝑔(𝒙 + ∆𝒙))
1
lim (
∙
)
∆𝑥→0 𝑔(𝑥)𝑔(𝒙 + ∆𝒙)
∆𝑥
1
𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥)
𝑔(𝑥 + ∆𝑥) − 𝑔(𝑥)
lim (
) ∙ ( lim 𝑔(𝑥)
− lim 𝑓(𝑥)
)
∆𝑥→0 𝑔(𝑥)𝑔(𝒙 + ∆𝒙)
∆𝑥→0
∆𝑥→0
∆𝑥
∆𝑥
lim (
∆𝑥→0
1
𝑓(𝑥 + ∆𝑥) − 𝑓(𝑥)
𝑔(𝑥 + ∆𝑥) − 𝑔(𝑥)
) ∙ ( lim 𝑔(𝑥) ∙ lim
− lim 𝑓(𝑥) ∙ lim
)
∆𝑥→0
∆𝑥→0
∆𝑥→0
∆𝑥→0
𝑔(𝑥)𝑔(𝒙 + ∆𝒙)
∆𝑥
∆𝑥
d [ ( )] ( ) d [ ( )] ( )
𝑓𝑥 𝑔𝑥 −
𝑔𝑥 𝑓𝑥
1
d
d
d𝑥
d𝑥
=
∙ (𝑔(𝑥) ∙ [𝑓(𝑥)] − 𝑓(𝑥) [𝑔(𝑥)]) =
[𝑔(𝑥)]2
d𝑥
d𝑥
𝑔2 (𝑥)
Sixth:
Find the derivative of the function using the limit process:
𝑓(𝑥) =
3
𝑥7
Solution: In this problem, we must use formula for finding derivative
of f(x)
𝑓(𝑥 + 𝛥𝑥) − 𝑓(𝑥)
𝛥𝑥
∆𝑥→0
𝑓 ′ (𝑥) = 𝑙𝑖𝑚
1
1
− 7
7
(𝑥 + 𝛥𝑥)
𝑥
𝑓 ′ (𝑥) = lim 3 ∙ 𝑙𝑖𝑚
=
𝛥𝑥
∆𝑥→0
∆𝑥→0
lim 3 ∙ lim
∆𝑥→0
∆𝑥→0
1
1
−
𝑥 7 + 7𝑥 6 ∆𝑥 + 21𝑥 5 ∆𝑥 2 + 35𝑥 4 ∆𝑥 3 + 35𝑥 3 ∆𝑥 4 + 21𝑥 2 ∆𝑥 5 + 7𝑥∆𝑥 6 + ∆𝑥 7 𝑥 7
∆𝑥
lim 3 ∙ lim (
∆𝑥→0
∆𝑥→0
𝑥 7 − 𝑥 7 − 7𝑥 6 ∆𝑥 − 21𝑥 5 ∆𝑥 2 − 35𝑥 4 ∆𝑥 3 − 35𝑥 3 ∆𝑥 4 − 21𝑥 2 ∆𝑥 5 − 7𝑥∆𝑥 6 − ∆𝑥 7
)
(𝑥 7 + 7𝑥 6 ∆𝑥 + 21𝑥 5 ∆𝑥 2 + 35𝑥 4 ∆𝑥 3 + 35𝑥 3 ∆𝑥 4 + 21𝑥 2 ∆𝑥 5 + 7𝑥∆𝑥 6 + ∆𝑥 7 ) ∙ 𝑥 7 ∙ ∆𝑥
lim 3 ∙ lim (
∆𝑥→0
∆𝑥→0
−7𝑥 6 − 21𝑥 5 ∆𝑥 − 35𝑥 4 ∆𝑥 2 − 35𝑥 3 ∆𝑥 3 − 21𝑥 2 ∆𝑥 4 − 7𝑥∆𝑥 5 − ∆𝑥 6
)
(𝑥 7 + 7𝑥 6 ∆𝑥 + 21𝑥 5 ∆𝑥 2 + 35𝑥 4 ∆𝑥 3 + 35𝑥 3 ∆𝑥 4 + 21𝑥 2 ∆𝑥 5 + 7𝑥∆𝑥 6 + ∆𝑥 7 ) ∙ 𝑥 7
Using direct substitution for solving this limit process:
lim 3 = 3
∆𝑥→0
−7𝑥 6
21
3 ∗ 14 = − 8
𝑥
𝑥
3
Answer: the derivative of 𝑓(𝑥) = 𝑥 7 is equal to −
21
𝑥8