MATH1510 Calculus for Engineers (Fall 2016)
Suggested Solution for Midterm Examination
1. Solution:
(a) Noted (1 + x)3 = x3 + 3x2 + 3x + 1,
lim−
x→0
(1 + x)3 − 1
x3 + 3x2 + 3x + 1 − 1
= lim−
x→0
x
x
2
= lim− (x + 3x + 3) = 3.
x→0
(b)
(x − 1)(x − 2)
x2 − 3x + 2
= lim
2
x→1 (1 − x)(1 + x)
x→1
1−x
2−x
1
= lim
= .
x→1 1 + x
2
lim
(c)
x5 x4 x3 x2
1
x 2 5+ 5+ 5+ 5+ 5
2x5 + x4 + x3 + x2 + 1
x
x
x
x
x
=
lim
lim
4
3
2
5
x→+∞
x→+∞ πx5 + x4 + x3 + x2 + 1
x
x
x
1
x
x5 π 5 + 5 + 5 + 5 + 5
x
x
x
x
x
−1
−2
−3
−5
2
2+x +x +x +x
= .
= lim
x→+∞ π + x−1 + x−2 + x−3 + x−5
π
5
(d)
x −1
x −1
2 2
2
1 2
lim 1 +
1+
= lim 1 +
x→+∞
x→+∞
x
x/2
x
x2
1
1
lim
= lim 1 +
x→+∞ 1 + 2
x→+∞
x/2
x
= e · 1 = e.
(e) Since | sin x| ≤ 1 and | cos3 x| ≤ 13 = 1, and as x → −∞,
2x
x(sin x + cos3 x)
2x
≤
≤
−
,
x2 + 1
x2 + 1
x2 + 1
moreover,
2
2x
= lim
= 0,
2
x→−∞ 1/x + x
x→−∞ 1 + x
lim
lim (−
x→−∞
then using the squeeze theorem, we know
x(sin x + cos3 x)
lim
= 0.
x→−∞
x2 + 1
2x
) = 0,
1 + x2
2. Solution:
(a) First, the right hand limit is,
lim f (x) = lim+ sin x = 0
x→0+
x→0
Second, the left hand side limit is,
lim f (x) = lim− −|x| = 0
x→0
x→0−
and by the definition of f (x), one have
f (0) = −|0| = 0.
The above three identities gives
lim f (x) = lim− f (x) = f (0) = 0
x→0+
x→0
by definition of continuity of a function at a point, we know f is continuous at
x = 0.
(b) By definition of Lf ′ and f (x) for x < 0,
Lf ′ (0) = lim−
h→0
−|h| − 0
−(−h)
f (h) − f (0)
= lim−
= lim−
= 1.
h→0
h→0
h
h
h
(c) By definition of Rf ′ and f (x) for x > 0,
Rf ′ (0) = lim+
h→0
f (h) − f (0)
sin h − 0
= lim+
= 1.
h→0
h
h
(d) By (b) and (c), we know
Lf ′ (0) = Rf ′ (0) = 1
so f (x) is differentiable at x = 0.
20
15
10
y
5
0
−5
−10
−15
−20
−20
−15
−10
−5
0
x
5
Figure 1: Question 2
10
15
20
3. Solution:
(a) For x > 0, noted
d a
x = axa−1 ,
dx
dy
2
3
= x−1/3 − x−5/2 .
dx
3
2
(b) For x > 0, noted log9 x =
d x
ln x
and
9 = 9x ln 9,
ln 9
dx
dy
2 1 9x
=
+
ln 9.
dx
ln 9 x
2
(c) Noted (f /g)′ = (f ′ g − f g ′)/g 2,
cos x(cos x + 3) + sin x(sin x + 2)
1 + 3 cos x + 2 sin x
dy
=
=
.
2
dx
(cos x + 3)
(cos x + 3)2
(d) By chain rule,
dy
= ecos x · (− sin x) + cos (ex ) · ex = −ecos x sin x + ex cos ex .
dx
(e) For x > 0, taking natural logarithms of both sides of y = x1/x
ln y =
1
ln x,
x
then taking derivative with respect to x,
1
1
1 dy
= − 2 ln x + 2 ,
y dx
x
x
then replace y by x1/x ,
dy
= −x1/x−2 ln x + x1/x−2 .
dx
4. Solution: By the representation of f (x),
√
f ′ (x) = −2 sin (2x) + 2 2 sin x = 4 sin x
√
!
2
− cos x
2
Take f ′ (x) = 0, one can get
sin x = 0 or cos x =
√
π
2
.
= cos ±
2
4
Case 1: sin x = 0, then x = kπ , k ∈ Z, where Z is the set whose elements are all
the integers( same notation also used in the following).
√
2
π
, then x = ± + 2kπ, k ∈ Z
Case 2: cos x =
2
4
Combine Case 1 and 2, one can get all the critical points of f (x) are
n
o n π
o
kπ, k ∈ Z ∪ ± + 2kπ, k ∈ Z .
4
10
y=f
y = f′
8
6
4
y
2
0
−2
−4
−6
−8
−10
−10
−5
0
x
Figure 2: Question 4
5
10
5. Solution: Firstly, we need to get the second derivative of f (x),
f ′ (x) = ex + xex ,
f ′′ (x) = ex + ex + xex = (2 + x)ex .
Note that ex is positive, so f ′′ (x) > 0 for x > −2; and f ′′ (x) < 0 for x < −2.
So the only inflection point is − 2, f (−2) = (−2, −2e−2 ) where the concavity
switches from negative to positive.
IP
x = −2
Test Value
x = −10
′′
Sign of f (x) f ′′ (x) < 0
−
x = 10
f (x) > 0
+
′′
35
y=f
y = f′
y = f′′
30
25
y
20
15
10
5
0
−3
−2.5
−2
−1.5
−1
−0.5
x
0
0.5
Figure 3: Question 5
1
1.5
2
6. Solution:
(a) Calculate the derivatives first,
y ′ = ex sin (2x) + ex · 2 cos (2x) = ex sin (2x) + 2 cos (2x)ex ,
y ′′ = ex sin (2x) + 2ex cos (2x) + 2 ex cos (2x) − 2ex sin (2x)
= ex − 3 sin (2x) + 4 cos (2x) .
Substitute the formulas of y, y ′ , y ′′ to the equation for y,
0 = y ′′ − 2y ′ + ky
= ex − 3 sin (2x) + 4 cos (2x) − 2[ex sin (2x) + 2 cos (2x)ex ] + kex sin(2x)
= (k − 5)ex sin (2x),
then
k = 5.
400
ky where k = 5
y′
y′′
300
200
100
y
0
−100
−200
−300
−400
−500
−5
−4
−3
−2
−1
0
x
1
2
Figure 4: Question 6a
3
4
5
(b) Taking derivative with respect to x,
x2
e
dy
dy
· 2x + 2y
= 0,
+ 2 cos (2xy) y + x
dx
dx
i.e
2[y + x cos (2xy)]
then
dy
2
= −2 y cos (2xy) + xex ,
dx
2
y cos (2xy) + xex
dy
=−
.
dx
y + x cos (2xy)
Evaluate at (x, y) = (0, 1), one get
dy 1+0
=−
= −1.
dx (0,1)
1+0
which gives the slope of the tangent line, and since the tangent line pass the
point (0, 1) , then the equation of the tangent line is:
y = −x + 1.
2
2
exp(x )+y +sin(2 x y)−2 = 0
2
y = f(x)
tangent line at (0, 1)
1.5
1
y
0.5
0
−0.5
−1
−1.5
−2
−5
−4
−3
−2
−1
0
x
1
2
Figure 5: Question 6b
3
4
5
(c)
1
x
11 Since f (x) is continuous and differentiable in
1, 10 , one can use Lagrange’s
11
mean value theorem, i.e, there is a c ∈ 1, 10 , such that
− f (1)
f 11
1
10
= f ′ (c) = ,
11
c
−1
10
f ′ (x) =
then noted f (1) = 0 and f (x) = ln x, one have
1 1
11
=
.
ln
10
10 c
By the fact
1
10
< < 1 for c ∈ (1, 11
),
10
11
c
1 1
1
1
<
< .
11
10 c
10
Combining the above two, one get the conclusion.
(d)
i.
Solution 1: By using l’Hôpital’s rule twice,
0
)
0
0
0
e−3r − e−2r + r
lim
r→0
r2
−3e−3r + 2e−2r + 1
= lim
r→0
2r
5
9e−3r − 4e−2r
= .
r→0
2
2
= lim
Solution 2: By the Taylor expansion of f (x) = ex at 0,
ex = 1 + x + x2 /2 + o(x2 )
then
e−3r − e−2r + r
(1 − 3r + 9r 2 /2) − (1 − 2r + 4r 2/2) + r + o(r 2 )
=
lim
r→0
r→0
r2
r2
5 o(r 2 )
5
5r 2 /2 + o(r 2 )
=
lim
+
=
.
= lim
r→0
r→0
r2
2
r2
2
lim
10
9
8
7
6
y
5
4
3
2
1
0
−1
−1
−0.8
−0.6
−0.4
−0.2
0
x
0.2
0.4
Figure 6: Question 6(d)i
0.6
0.8
1
ii. First,
ln(aK
−t
−t −1/t
+ (1 − a)L )
0
.
0
ln(aK −t + (1 − a)L−t )
=−
t
where since lim A−t = 1 for A > 0, then
t→0
lim ln(aK −t + (1 − a)L−t ) = ln(a + (1 − a)) = ln 1 = 0.
t→0
By using l’Hôpital’s rule,
lim −
t→0
ln(aK −t + (1 − a)L−t )
aK −t ln K + (1 − a)L−t ln L
= lim
t→0
t
aK −t + (1 − a)L−t
= a ln K + (1 − a) ln L = ln(K a L1−a ).
a = 0.3 K = 5 L = 9
2.1
2.05
y
(0, ln(50.3 90.7 )
2
1.95
1.9
1.85
−3
−2
−1
0
x
1
Figure 7: Question 6(d)ii
2
3
iii. First, we consider
lim (2u − π) ln(tan u)
u→ π2 −
ln(tan u)
(2u − π)−1
sec2 u
2
−
= lim−
tan u
(2u − π)2
u→ π2
0
( )
0
= lim−
u→ π2
= − lim
π−
u→ 2
= lim
π−
u→ 2
(2u − π)2
(2u − π)2
= lim
sin (2u)
u→ π2 − sin (2u − π)
(2u − π)
· (2u − π) = 0.
sin (2u − π)
x
= 1, so
x→0 sin x
here we used l’Hôpital’s rule and lim
lim (tan u)2u−π = e0 = 1.
u→ π2 −
10
y
8
6
4
( π2 , 1)
y
2
0
−2
−4
−6
−8
−10
1
1.1
1.2
1.3
1.4
1.5
x
1.6
1.7
Figure 8: Question 6(d)iii
iv.
1.8
1.9
2
Solution 1: By l’Hôpital’s rule
sin−1 (x/6)
lim
x→0 tan−1 (x/2)
1
1
p
6 1 − x2 /36
1
= .
= lim
1
1
x→0
3
2
2 1 + x /4
0
0
Solution 2: As sin−1 (x) = x + o(x), tan−1 x = x + o(x), then
sin−1 (x/6)
1
x/6 + o(x)
1/6 + o(x)/x
= lim
= lim
= .
−1
x→0 tan (x/2)
x→0 x/2 + o(x)
x→0 1/2 + o(x)/x
3
lim
0.5
y
0.45
0.4
(0, 13 )
0.35
y
0.3
0.25
0.2
0.15
0.1
0.05
0
−1
−0.8
−0.6
−0.4
−0.2
0
x
0.2
0.4
Figure 9: Question 6(d)iv
0.6
0.8
1