MEMORIAL UNIVERSITY OF NEWFOUNDLAND
DEPARTMENT OF MATHEMATICS AND STATISTICS
Test 1
MATH 1000
Fall 2015
SOLUTIONS
[12]
1. (a) f (0) = 2
(b) lim− f (x) = −2
x→0
(c) lim+ f (x) = 2
x→0
(d) lim f (x) does not exist (because the one-sided limits are not equal)
x→0
(e) f (−3) = −4
(f)
(g)
lim f (x) = 0
x→−3−
lim f (x) = 0
x→−3+
(h) lim f (x) = 0
x→−3
(i) f (2) is undefined
(j) lim− f (x) = −∞
x→2
(k) lim+ f (x) = −∞
x→2
(l) lim f (x) = −∞ (and therefore does not exist)
x→2
[6]
2. (a) Direct substitution yields a difference of K0 but, since the resulting limits would not
exist, we cannot rewrite this as the difference of two limits. Instead, we rewrite the given
function as a single rational function, and then use the Cancellation Method:
1
1
6
6
− 2
−
lim
= lim
x→−3 x2 − 9
x→−3 (x − 3)(x + 3)
x + 5x + 6
(x + 2)(x + 3)
6(x + 2) − (x − 3)
(x − 3)(x + 2)(x + 3)
5x + 15
= lim
x→−3 (x − 3)(x + 2)(x + 3)
= lim
x→−3
5(x + 3)
x→−3 (x − 3)(x + 2)(x + 3)
5
= lim
x→−3 (x − 3)(x + 2)
5
=
−6 · (−1)
5
= .
6
= lim
[6]
(b) Direct substitution results in a 00 indeterminate form. Since this is a quasirational function, we use the Rationalisation Method:
√
√
(3x − 6) 3 + 5x − 1
3 + 5x − 1
3x − 6
√
√
·
= lim
lim
x→2 3 −
9 − (5x − 1)
5x − 1 3 + 5x − 1 x→2
√
3(x − 2) 3 + 5x − 1
= lim
x→2
10 − 5x
√
3(x − 2) 3 + 5x − 1
= lim
x→2
−5(x − 2)
√
3 3 + 5x − 1
= lim
x→2
−5
3(3 + 3)
=
−5
18
=− .
5
[6]
(c) Again, we obtain a
limit
0
0
indeterminate form by direct substitution. But recall the special
sin(t)
= 1.
t→0
t
lim
We can rewrite the given limit as
sin(4x) cos(5x)
1
sin(4x) 4x cos(5x)
= lim
·
x→0
6 sin(7x)
6 x→0 4x
sin(7x)
lim
=
sin(4x)
x
1
lim
· lim
· 4 cos(5x)
x→0 sin(7x)
6 x→0 4x
=
1
sin(4x)
7x
4 cos(5x)
lim
· lim
·
x→0 sin(7x)
6 x→0 4x
7
1
sin(4x)
7x
4
lim
· lim
· lim cos(x)
x→0 sin(7x) 7 x→0
6 x→0 4x
1
4
= · 1 · 1 · · cos(0)
6
7
2
=
·1
21
2
= .
21
=
[6]
3. For the limit to exist, the one-sided limits must be equal. From the left,
lim f (x) = lim− (3x2 − 5) = 3p2 − 5.
x→p−
x→p
From the right,
lim f (x) = lim+ = p2 + 9p.
x→p+
x→p
Setting these equal to each other, we have
3p2 − 5 = p2 + 9p
2p2 − 9p − 5 = 0
(2p + 1)(p − 5) = 0,
so p = − 12 or p = 5.
[4]
4. (a) This statement is false. Observe that
f (x) =
(x − 4)(x + 4)
x2 − 16
=
= x − 4,
x+4
x+4
but this cancellation is valid only if x 6= −4. Thus the two functions agree everywhere
except at x = −4.
(Alternatively, we might observe that if we substitute x = −4 directly into f (x), we
obtain division by zero, which indicates that f (−4) is undefined. However, g(−4) = −8.)
(b) This statement is true. Because f (x) = g(x) for all x 6= −4, and the behaviour of f (x)
and g(x) at x = −4 does not affect their limits as x → −4, lim f (x) = lim g(x).
x→−4
x→−4