Sample Answer Sheet 3
MATH10131
Limits and Derivatives
Easy Questions
1.
(a)
−6x(1 − x2 )2
(e)
3u2 + 1
√
2 u3 + u
(b)
2 sin t cos t
1
(1 − x)2
(c)
− e−t cosh t sinh t
(f)
−t
+e
−t
+e
2
sinh t
3x2 exp(−2x)
(g)
−2x3 exp(−2x)
2
cosh t
(d)
− 2z −3 ln(1 + z 2 )
(h)
n o
sin(2x)
2 cos(2x)
0
=
=2
= lim
x
0
1
x→0
n
o
n
o
ln(1 + ex )
ln 1
0
(b) lim
=
=
=0
x→−∞
x
−∞
−∞
n o
sinh(x)
cosh(x)
0
=
= lim
=1
(c) lim
x
0
1
x→0
x→0
2. (a) lim
x→0
(
2
e−1/x
3. (a) f (x) =
A
( √
1/ x4
(b) g(x) =
A
©
ª
2
lim e−1/x = e−∞ = 0. So choose A = 0.
if x 6= 0
if x = 0.
x→0
if x 6= 0
if x = 0.
n o
1
1
= lim 2 =
x→0 |x|
x4
1
0
lim √
x→0
= ∞.
So no choice of value of A can make g(x) continuous.
Standard Questions
4. (a)
(b)
d
dx
d
du
d
da
sin2 x cos3 x = 2 sin x cos4 x − 3 sin3 x cos2 x
(au + b)3/2 = 32 a(au + b)1/2
(au + b)3/2 = 32 u(au + b)1/2
r
r
a − 6 x + (a + 6 x)
a+x
1 a−x
d
(d)
=
×
=
2
(c)
a−x
dx
2
a+x
(a − x)
a
(a − x)2
r
a−x
a+x
s
s
cos(s/p) = sin(s/p) − cos(s/p)
p62
p
(e)
d
dp
p sin(s/p) = sin(s/p)− 6 p
(f)
d
dθ
θn sinm (aθ + b) = n θn−1 sinm (aθ + b) + ma θn sinm−1 (aθ + b) cos(aθ + b)
5. (a) Consecutive derivatives of
1
−1
2
−2 × 3 2 × 3 × 4
are:
,
,
,
, ···
1+x
(1 + x)2 (1 + x)3 (1 + x)4 (1 + x)5
(−1)n n!
dn
1
=
n
dx 1 + x
(1 + x)n+1
1
1
2
2×3
2×3×4
(b) Consecutive derivatives of
are:
,
,
,
, ···
1−x
(1 + x)2 (1 + x)3 (1 + x)4 (1 + x)5
So we can write:
So we can write:
(
tt
6. (a) g(t) =
A
dn
1
n!
=
dxn 1 + x
(1 + x)n+1
©
ª
lim+ tt = lim+ et ln t = e0×(−∞) .
t→0
t→0
n
o
1/t
−∞
t
ln t
=
= lim+
= lim+
= 0.
lim+ t ln t = lim+
2
t→0
Hence
if t > 0
if t = 0.
t→0
lim e
t→0+
t ln t
1/t
∞
t→0
−1/t
0
t→0
−1
= e = 1. So setting A = 1 makes g(t) continuous on [0, ∞).
1
2θ tan θ + θ2 sec2 θ
+ z −2
2z
1 + z2
7. Are the following true or false?
1
1
1
= 0. This is false because lim+
= 0 and lim−
= 1.
x→0 1 + e1/x
x→0 1 + e1/x
1 + e1/x
The two-sided limit does not exist.
©
ª
1
π
1
(b) lim tan−1
= . This is false because lim tan−1
= tan−1 ∞ = π2 and
−
x→1
1−x
2
1−x
x→1
©
ª
1
lim tan−1
= tan−1 (−∞) = − π2 . The two-sided limit does not exist.
1−x
x→1+
(
1
for x ≥ 0
8. f (x) = 1 + x2 and g(x) =
−1 for x < 0.
(
)
¡
¢
¡
¢
1 + 12
for x ≥ 0
f g(x) =
= 2 for all x ∈ R, so that f g(x) is continuous
2
1 + (−1) for x < 0
¡
¢
¡
¢
f (x) = 1 + x2 > 0 for all x ∈ R so g f (x) = 1 for all x, so that g f (x) is continuous.
(a) lim
x→0
Harder Questions
9. (a) x − [x] is discontinuous at every integer x ∈ Z
(b) [sin x] is discontinuous where sin x = 0 (i.e. x = kπ for k ∈ Z)
and where sin x = 1 (i.e. x = π2 + 2kπ for k ∈ Z).
It is not discontinuous where sin x = −1 (why?).
(c) [1/x] is discontinuous where 1/x ∈ Z and at x = 0.
(d) (−1)[x] is discontinuous at every integer x ∈ Z, where the value swops between −1 and 1.
¡
¢[ln x2 ]
2
2
(e) (−1)2[ln x ] = (−1)2
= 1[ln x ] is only discontinuous at x = 0 where it is not defined.
2