NATIONAL UNIVERSITY OF SINGAPORE
MA2002 Calculus
Solution to Tutorial 1
T UTORIAL PART I
1. The domain of f is R and the domain of g is {x | x != 0} = R \ {0}.
(i) f ◦ g (x) = f (g (x)) = f (1/x) = 1 − 1/x 3 , and the domain of f ◦ g is given by
{x | x != 0 and 1/x ∈ R} = R \ {0}.
(ii) g ◦ f (x) = g ( f (x)) = g (1 − x 3 ) = 1/(1 − x 3 ), and the domain of g ◦ f is given by
{x | x ∈ R and 1 − x 3 != 0} = R \ {1}.
(iii) f ◦ f (x) = f ( f (x)) = f (1 − x 3 ) = 1 − (1 − x 3 )3 , and the domain of f ◦ f is given by
{x | x ∈ R and 1 − x 3 ∈ R} = R.
(iv) g ◦ g (x) = g (g (x)) = g (1/x) = 1/(1/x) = x, and the domain of g ◦ g is given by
{x | x != 0 and 1/x != 0} = R \ {0}.
%
%
2. g ◦ h(x) = g (h(x)) = g ( x + 3) = cos x + 3, and
%
f ◦ g ◦ h(x) = f (g ◦ h(x)) = f (cos x + 3) =
%
2
cos x + 3 + 1
.
3. (a) f (x) = x −3 is odd, for f (−x) = (−x)−3 = −(x −3 ) = − f (x).
(b) f (x) = |sin x| − 4x 3 is even, for f (−x) = |sin(−x)| − 4(−x)2 = |sin x| − 4x 2 = f (x).
(c) f (x) = 3x 3 + 2x 2 + 1 is neither odd nor even. For instance, f (1) = 6 and f (−1) = 0.
4. (a) lim (x + 3)2021 = (−4 + 3)2021 = (−1)2021 = −1.
x→−4
u4 − 1
(u − 1)(u 3 + u 2 + u + 1)
u 3 + u 2 + u + 1 13 + 12 + 1 + 1 4
=
lim
=
lim
=
= .
u→1 u 3 − 1
u→1
u→1
(u − 1)(u 2 + u + 1)
u2 + u + 1
12 + 1 + 1
3
%
%
x −1
(x − 1)( x + 3 + 2)
(x − 1)( x + 3 + 2)
(c) lim %
= lim %
= lim
%
x→1 x + 3 − 2
x→1 ( x + 3 − 2)( x + 3 + 2)
x→1
(x + 3) − 22
%
%
= lim ( x + 3 + 2) = 1 + 3 + 2 = 4.
(b) lim
x→1
1
MA2002 CALCULUS TUTORIAL SOLUTION 1
2
x 2 − 4x
x(x − 4)
x
4
4
= lim
= lim
=
= .
2
x→4 x − 3x − 4
x→4 (x + 1)(x − 4)
x→4 x + 1
4+1 5
(d) lim
x 2 − 4x
(e) lim 2
does not exist, because
x→−1 x − 3x − 4
lim (x 2 − 3x − 4) = (−1)2 − 3(−1) − 4 = 0,
x→−1
but
lim (x 2 − 4x) = (−1)2 − 4(−1) = 5 != 0.
x→−1
f (x)
In general, if lim
exists and equals L ∈ R, and lim g (x) = 0, we must have
x→a g (x)
x→a
f (x)
· lim g (x) = L · 0 = 0.
x→a
x→a g (x) x→a
%
%
%
1+h −1
( 1 + h − 1)( 1 + h + 1)
(1 + h) − 12
(f) lim
= lim
= lim %
%
h→0
h→0
h→0 h( 1 + h + 1)
h
h( 1 + h + 1)
1
1
1
= lim %
=%
= .
h→0 1 + h + 1
1+0+1 2
lim f (x) = lim
(g) Note that for all x > 0, −1 ≤ sin(1/x) ≤ 1. Then
%
%
%
− x ≤ x sin(1/x) ≤ x.
%
%
%
Since lim+ (− x) = lim+ x = 0, by squeeze theorem lim+ x sin(1/x) exists and equals
0.
x→0
x→0
x→0
%
%
%
%
%
x − x2
( x − x 2 )(1 + x)
x + x − x2 − x2 x
(h) lim
= lim
% = lim
%
%
x→1 1 − x
x→1 (1 − x)(1 + x)
x→1
1−x
%
%
%
x − x 2 + x (1 − x 2 )
= lim
= lim (x + x (1 + x)) = 1 + 1(1 + 1) = 3.
x→1
x→1
1−x
!
!
!1
!
1
1
10
10
5. !! − 0.5!! < 0.2 ⇔ −0.2 < − 0.5 < 0.2 ⇔ 0.3 < < 0.7 ⇔
<x< .
x
x
x
7
3
So we shall find a proper δ > 0 such that (2−δ, 2+δ)\{2} is a subset of the interval (10/7, 10/3).
10
7
2−δ
!"
2
2+δ
10
3
x
It follows that 10/7 ≤ 2 − δ and 2 + δ ≤ 10/3. We have δ ≤ 4/7 and δ ≤ 4/3. So δ ≤ 4/7. In
particular, we may take δ = 4/7 (or any positive number ≤ 4/7).
f (x)
exists and lim g (x) = 0, then lim f (x) = 0.
x→a g (x)
x→a
x→a
2
2
ax + a x − 2
Suppose lim 3
exists. Since lim (x 3 − 3x + 2) = 13 − 3 · 1 + 2 = 0, we must have
x→1 x − 3x + 2
x→1
lim (ax 2 + a 2 x − 2) = 0, that is, a + a 2 − 2 = 0. Therefore, a = 1 or −2.
6. Recall the result in 4(e): If lim
x→1
MA2002 CALCULUS TUTORIAL SOLUTION 1
3
ax 2 + a 2 x − 2
x2 + x − 2
(x − 1)(x + 2)
1
=
lim
= lim
= lim
which does
3
3
2
x→1 x − 3x + 2
x→1 x − 3x + 2
x→1 (x − 1) (x + 2)
x→1 x − 1
If a = 1, then lim
not exist.
ax 2 + a 2 x − 2
−2x 2 + 4x − 2
−2(x − 1)2
=
lim
=
lim
x→1 x 3 − 3x + 2
x→1 x 3 − 3x + 2
x→1 (x − 1)2 (x + 2)
−2
−2
2
= lim
=
=− .
x→1 x + 2
1+2
3
If a = −2, then lim
7. If x is rational, then 0 ≤ f (x) = x 2 ; if x is irrational, then 0 = f (x) ≤ x 2 . So for any real number
x, we have 0 ≤ f (x) ≤ x 2 .
Since lim 0 = lim x 2 = 0, by squeeze theorem lim f (x) exists and equals 0.
x→0
x→0
x→0
T UTORIAL PART II
1. Observe that inductively,
2
n
f n (x) = f 0 ( f n−1 (x)) = f n−1 (x)2 = ( f n−2 (x)2 )2 = f n−2 (x)2 = · · · = f 0 (x)2 .
n
This holds for n = 1, 2, . . . . Since f 0 (x) = x 2 , we have f n (x) = (x 2 )2 = x 2
n+1
.
#
"
#
2π
2π
2
2. For x != 1, as 0 ≤ cos
≤ 1, we have 4 ≤ 4 + cos
≤ 5 and so
x −1
x −1
"
"
##
2π
2
2
2
4(x − 1) ≤ (x − 1) 4 + cos
≤ 5(x − 1)2 .
x −1
2
"
Since lim 4(x − 1)2 = lim 5(x − 1)2 = 0, by Squeeze Theorem,
x→1
x→1
"
"
##
2π
2
2
lim (x − 1) 4 + cos
= 0.
x→1
x −1
|x|
3. For example, let f (x) = g (x) =
=
x
$
1, if x > 0,
−1, if x < 0.
|x|
|x|
|x|
= 1 and lim−
= −1 are not equal, lim
does not exist.
x→0 x
x→0 x
x→0 x
"
#
|x| |x|
|x| |x|
On the other hand,
= 1 for all x ∈ R\{0}, so lim
= lim 1 = 1.
x→0 x x
x→0
x x
Since lim+
4.
i) Find the coordinates of Q:
$
(x − 1)2 + y 2 = 1,
r2
.
2
x 2 + y 2 = r 2,
This is the x-coordinate of the intersection Q. So the y-coordinates of Q is
&
' r (2
%
2
2
r −x = r 1−
.
2
Solving simultaneous equations
we have x =
MA2002 CALCULUS TUTORIAL SOLUTION 1
4
ii) Find the equation of PQ:
The equation of the straight line passing through P (0, r ) and Q is given by
)
)
r 2
r
1
−
(
)
−
r
1 − ( r2 )2 − 1
y −r
2
=
=
.
r
r2
x −0
−
0
2
2
iii) Find the x-coordinate of R:
Note that R is on the x-axis. Let y = 0 in the equation of PQ. Then the x-coordinate of R
r2
is given by x = '
(.
)
2 1 − 1 − ( r2 )2
iv) Find the limiting position of R:
We can evaluate the limit
'
(
)
r 2
2
r
1
+
1
−
(
)
r
2
lim+ '
( = lim+ *
'
(
)
)
)
+
r →0
r →0
2 1 − 1 − ( r2 )2
2 1 − 1 − ( r2 )2 1 + 1 − ( r2 )2
'
(
)
r 2 1 + 1 − ( r2 )2
= lim+
r →0
2 · ( r2 )2
&
"
#
* r +2
= lim+ 2 1 + 1 −
r →0
2
2
= 4.
Therefore, as r → 0+ , R → (4, 0) along the x-axis.