The University of Sydney
School of Mathematics and Statistics
Solutions to Board Tutorial 4
MATH1021: CALCULUS OF ONE VARIABLE
LIMITS AND CONTINUITY
1. Complete the tables below and use them to guess the value of lim
√
x→16
Evaluate f (x) correct to five decimal places.
x < 16
15
15.5
15.9
15.95
15.99
17
16.5
16.1
16.05
16.01
x−4
.
x − 16
15.999
f (x)
x > 16
16.001
f (x)
IMPORTANT The aim of this problem is to help you become familiar with the
concept of lim f (x) in an informal way, particularly the idea that the point a itself
x→a
is not relevant as far as limits are concerned and that only nearby points are of
consequence.
As a matter of fact, if we substitute x = 16 into f (x) we find that the function is
undefined at that point, however when we complete the tables, we will see that the
values of the function approach a definite value when x gets closer to 16.
In practice we do not have to construct tables. Instead, we use the Basic Limit
Laws and other important properties introduced in lectures. In a later chapter
we will study L’Hôpital’s rule, which is a powerful technique for finding limits of
indeterminate forms.
√
x−4
Solution:
lim
. If we substitute x = 16 into the function, we get
x→16 x − 16
0
an indeterminate form of type . Completing the first table we start approaching
0
x = 16 from the left and we notice that the function approaches the value y = 0.125.
That is, lim− f (x) = 0.125
x→16
x < 16
15
15.5
15.9
15.95
f (x)
0.12701
0.12599
0.12519
0.12509
15.99
15.999
0.12501 0.12500
Completing the second table, we approach x = 16 from the right and the limit
approaches the same value y = 0.125. That is, lim+ f (x) = 0.125.
x→16
Copyright c 2021 The University of Sydney
1
Since lim− f (x) = lim+ f (x) = 0.125, we guess that lim f (x) = 0.125.
x→16
x→16
x→16
x > 16
17
16.5
16.1
16.05
16.01
f (x)
0.12310
0.12403
0.12480
0.12490
16.001
0.12498 0.12499
We show in Exercise 2 that the exact value is indeed lim f (x) =
x→16
1
= 0.125.
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2. Find the following limits by cancelling out the common factor in the numerator
and denominator:
x2 − 2x
(a) lim 2
x→2 x − x − 2
Solution: The numerator and denominator are both zero when x = 2 and
hence have x − 2 as a factor. So
lim
x→2
x2 − 2x
x(x − 2)
x
2
2
= lim
= lim
=
= ,
2
x − x − 2 x→2 (x + 1)(x − 2) x→2 x + 1
2+1
3
for all x such that x 6= 2 and x 6= −1.
x2 − 6x + 8
(b) lim 2
x→2 x − 5x + 6
Solution: The numerator and denominator are both zero when x = 2 and
hence have x − 2 as a factor. So
lim
x→2
2−4
x2 − 6x + 8
(x − 4)(x − 2)
(x − 4)
=
lim
=
lim
=
= 2,
x2 − 5x + 6 x→2 (x − 3)(x − 2) x→2 (x − 3)
2−3
for all x such that x 6= 2 and x 6= 3.
√
x−4
(c) lim
(Note that this limit is the same as that in Question 1).
x→16 x − 16
Solution: The numerator
√ and denominator are both zero when x = 16.
However, because of the √ x term, the common √factor on
√ both, top and
bottom is not x − 16 but x − 4. Since x − 16 = ( x − 4)( x + 4), we have
√
√
x−4
x−4
1
1
1
√
lim
= lim √
= .
= lim √
=
x→16 x − 16
x→16 ( x − 4)( x + 4)
x→16
4+4
8
x+4
for all x such that x 6= 4 and x 6= 16.
x2 + 3x + 2
(d) lim
x→−1 3x2 + 10x + 7
Solution: The numerator and denominator are both zero when x = −1,
and hence have x + 1 as a factor. Indeed, x2 + 3x + 2 = (x + 1)(x + 2) and
3x2 + 10x + 7 = (x + 1)(3x + 7). So
x+2
x2 + 3x + 2
=
2
3x + 10x + 7
3x + 7
for all x such that x 6= −1 and x 6= 3. It follows that
lim
x→−1
−1 + 2
1
x2 + 2x + 3
x+2
= lim
=
= .
2
3x + 10x + 7 x→−1 3x + 7
−3 + 7
4
2
x3 − 3x + 2
x→1 x2 − 2x + 1
Solution: The numerator and denominator are each divisible by x − 1,
and
x3 − 3x + 2
(x2 + x − 2)(x − 1)
x2 + x − 2
=
=
x2 − 2x + 1
(x − 1)2
x−1
(e) lim
for all x 6= 1. Since x2 + x − 2 is also divisible by x − 1, we find that
x2 + x − 2
(x + 2)(x − 1)
x3 − 3x + 2
=
=
= x+2
2
x − 2x + 1
x−1
x−1
for all x 6= 1. Thus
lim
x→1
x3 − 3x + 2
= lim (x + 2) = 3.
x2 − 2x + 1 x→1
3. Find the limit as x → ∞ of the following rational functions by dividing top and
bottom by the largest power of the bottom.
2x − 3
(a) lim
x→∞ 4x + 5
Solution: Divide both numerator and denominator by the largest power
of x in the denominator (which is just x):
lim
2
−
(3/x)
2x − 3
2
1
2−0
2 − (3/x)
=
lim
= lim
= x→∞
= = .
x→∞ 4x + 5
x→∞ 4 + (5/x)
4+0
4
2
lim 4 + (5/x)
x→∞
2
5+x
x→∞ 3 + 27x − x3
Solution: Divide numerator and denominator by x3 (highest power in
denominator):
(b) lim
5 + x2
0+0
(5/x3 ) + (1/x)
=
lim
=
= 0.
3
3
2
x→∞ 3 + 27x − x
x→∞ (3/x ) + (27/x ) − 1
0+0−1
3x + 6
(c) lim
x→∞ 2x + 1
Solution: Divide both numerator and denominator by x:
lim 3 + (6/x)
3x + 6
3
3+0
3 + (6/x)
=
lim
= lim
= x→∞
= .
x→∞ 2x + 1
x→∞ 2 + (1/x)
2+0
2
lim 2 + (1/x)
lim
x→∞
2
6 + 7x
2 − 31x + 3x3
Solution: Divide numerator and denominator by x3 (highest power in
denominator):
(d) lim
x→∞
lim
x→∞
6 + 7x2
0+0
(6/x3 ) + (7/x)
=
lim
=
= 0.
3
3
2
x→∞ (2/x ) − (31/x ) − 3
2 − 31x + 3x
0+0−3
4. Use the squeeze law to find the following limits:
3
1
(a) lim x4 cos
x→0
x
Solution: Since −1 ≤ cos(1/x) ≤ 1 (for x 6= 0) and x4 ≥ 0 it follows that
−x4 ≤ x4 cos(1/x) ≤ x4 ,
for all x 6= 0. Since lim x4 = 0, the squeeze law implies that lim x4 cos(1/x)
x→0
x→0
exists and is equal to zero.
1
(b) lim sin x sin
x→0
x
Solution: Observe that if x 6= 0 then sin x sin(1/x) = | sin x| | sin(1/x)| ≤
| sin x|. Therefore,
−| sin x| ≤ sin x sin(1/x) ≤ | sin x|,
for all x 6= 0. (Note that we need the absolute value signs here because sin x
is sometimes positive and sometimes negative.) Since lim | sin x| = 0, the
x→0
squeeze law implies that lim sin x sin(1/x) exists and equals zero.
x→0
5. For each of the functions given by the formulas below, sketch the graph and decide
whether or not the function has a limit as x → 1. [Note: Only the informal idea
of the limit of a function is required here.]
(
1
if x < 1,
(a) f (x) =
−x if x > 1.
2
1
bc
Solution:
−2
−1
−1
1
2
bc
−2
If there is a limit, it must be a single number ℓ with the property that values
of f (x) can be made as close as we like to ℓ by choosing x sufficiently close
to 1. The graph shows that f (x) = 1 whenever x is less than 1, and so if the
limit exists it will have to be 1.
However, when x is close to 1 but greater than 1, the values of f (x) are close
to −1; in fact, the amount by which x exceeds 1 equals the amount by which
f (x) is less than −1. So the limit, if it existed, would have to be −1. Thus
there is no single ℓ that satisfies the requirements of the informal definition;
hence the
( function does not have a limit as x → 1.
x
if x < 1,
(b) f (x) =
2 − x if x > 1.
Solution:
1
bc
−1
−1
1
4
2
For this function the limit exists and is equal to 1. We can make f (x) as
close to 1 as we like by choosing x sufficiently close to 1 (but not equal to
1). Although not required for the answer, the following sample calculations
may help you to understand the process.
If, for example, we want to ensure that f (x) is within 0.01 of 1 (meaning
that 0.99 < f (x) < 1.01) then requiring x to be within 0.01 of 1 (that is,
0.99 < x < 1 or 1 < x < 1.01) will do it.
Similarly, we can guarantee that f (x) is within 0.001 of 1 (that is, 0.999 <
f (x) < 1.001) by requiring that x is within 0.001 of 1 (that is, 0.999 < x < 1
or 1 < x < 1.001). Clearly the same kind of thing will work with any
number, no matter how small, in place of 0.001.
(
x2 − 1 if x < 1,
(c) f (x) = √
x − 1 if x > 1.
2
1
Solution:
bc
−2
−1
−1
1
2
−2
For this function the limit exists and is equal to 0. Although not required
for the answer, the following sample calculation may help you to understand
the process. Suppose that we want f (x) to be within 0.01 of 0, that is,
−0.01 < f (x) < 0.01. For values of x less than 1 this can be achieved if
insist x that 0.995 < x < 1. (Check this for a few values of x, using your
calculator.)
For values of x greater than 1, it can be achieved by requiring that 1 < x <
1.0001. Combining these two observations shows that if x is within 0.0001
units of 1 on either side of 1, so that 0.9999 < x < 1 or 1 < x < 1.0001, then
f (x) is within 0.01 of 0.
6. Use continuity and the basic limit laws to evaluate the limits below. Recall that a
function f (x) is said to be continuous at a point a if lim f (x) = f (a).
x→a
√
3+ x
(a) lim √
x→4
3+x
Solution: We know that the ratio r(x) = f (x)/g(x) is continuous on
an interval [a, b] if f (x) and g(x)√are continuous on [a, b] and g(x) 6= 0 if
3+ x
is continuous if x ≥ 0 (because of the
x ∈ [a, b]. The ratio r(x) = √
3+x
√
√
x term) and the denominator is zero when x = − 3 which is negative.
5
Therefore, r(x) is continuous at x = 4 and so
√
√
3+ 4
3+2
3+ x
lim √
= r(4) = √
=√
≈ 0.872.
x→4
3+x
3+4
3+4
(b)
lim (x cos2 x)
x→π/4
Solution: The function cos x is continuous everywhere and so is cos2 x,
therefore the product x cos2 x is also continuous for all x ∈ R. This means
that it is also continuous at x = π/4 and hence
lim (x cos2 x) = (π/4) cos2 (π/4) = π/8 ≈ 0.3927.
x→π/4
7. Use continuity and the basic limit laws to evaluate the limits below:
(a) lim (x3 + 2x2 + x + 1)
x→2
Solution:
lim f (x) = lim (x3 + 2x2 + x + 1)
x→2
x→2
= lim x3 + [lim 2] × [lim x2 ] + lim x + lim 1 (sum and product laws)
x→2
x→2
x→2
x→2
x→2
= [lim x]3 + lim 2 × [lim x]2 + lim x + 1
x→2
3
x→2
x→2
x→2
(power law)
= 2 + 2 × 22 + 2 + 1
= 19.
2x2 + 1
x→1 x2 + x + 2
Solution:
(b) lim
2x2 + 1
x→1 x2 + x + 2
limx→1(2x2 + 1)
=
limx→1 (x2 + x + 2)
[limx→1 2] × [limx→1 x2 ] + limx→1 1
=
limx→1 x2 + limx→1 x + limx→1 2
[limx→1 2] × [limx→1 x]2 + limx→1 1
=
[limx→1 x]2 + limx→1 x + limx→1 2
2 × 12 + 1
= 2
1 +1+2
3
= .
4
lim f (x) = lim
x→1
(c)
lim−
x→4
√
16 − x2
6
(quotient law)
(sum and product laws)
(power law)
Solution:
√
lim− f (x) = lim− 16 − x2
x→4
x→4
q
=
lim (16 − x2 )
x→4−
q
lim 16 − [ lim− x]2
=
x→4−
x→4
√
√
2
= 16 − 4 = 0 = 0.
(root law)
(difference and power laws)
(d) lim ecos x + sin(ln x).
x→2
Solution:
lim f (x) = lim ecos x + sin(ln x)
x→2
x→2
cos[limx→2 x]
=e
+ sin(ln[lim x])
x→2
(sum and composition laws)
= ecos(2) + sin(ln(2))
≈ 1.298.
8. Which of the following functions f have removable discontinuities at the point
x = a? If the discontinuity is removable, find a function g that agrees with f for
x 6= a and is continuous at x = a.
x4 − 1
, a=1
(a) f (x) =
x−1
Solution: Substituting x = 1 into f (x) gives an indeterminate form of
0
and therefore f (x) is discontinuous at x = 1. However, using the
type
0
process of long division, we can write
lim
x→1
x4 − 1
(x − 1)(x3 + x2 + x + 1)
= lim
x→1
x−1
x−1
3
2
= lim (x + x + x + 1)
x→1
3
= 1 + 12 + 1 + 1 = 4.
Since the limit as x → 1 exists and is finite, the discontinuity is removable
by defining f (1) = 4. The function g that agrees with f for x 6= 1 and is
continuous at x = 1 is given by
 4
x − 1
if x 6= 1,
g(x) = x − 1

4
if x = 1.
x3 − x2 − 2x
, a=2
x−2
Solution: Substituting x = 2 into f (x) gives another indeterminate form
0
of type and therefore f (x) is discontinuous at x = 2. However, since the
0
(b) f (x) =
7
numerator can be factorized, it is possible to write
lim
x→2
x3 − x2 − 2x
x(x + 1)(x − 2)
= lim
x→2
x−2
x−2
= lim x(x + 1)
x→2
= 2(2 + 1) = 6.
Since the limit as x → 2 exists and is finite, the discontinuity is removable
by defining f (2) = 6. The function g that agrees with f for x 6= 2 and is
continuous at x = 2 is given by
 3
2
 x − x − 2x
if x 6= 2,
g(x) =
x−2

6
if x = 2.
Solution:
9. Find a constant c so that g is continuous on the domain (−∞, ∞), where g is the
function defined below:
(
x2 − c2 if x < 4,
g(x) =
cx + 20 if x ≥ 4.
Solution: The function g is continuous on R = (−∞, ∞) if and only if lim g(x) =
x→a
g(a) for all a ∈ R. Now the functions defined by the formulas x2 − c2 and cx + 20
are continuous everywhere, no matter what the value of c may be. So the only
possible value of a for which lim g(x) = g(a) could fail is a = 4. Hence our task is
x→a
to find a value of c such that lim g(x) = g(4).
x→4
Now limx→4 g(x) = ℓ holds if and only if limx→4− g(x) = ℓ and limx→4+ g(x) = ℓ
both hold, where the notations x → 4− and x → 4+ indicate that x approaches 4
from the left and the right, respectively. (The definition of limx→4− g(x) is the
same as the definition of limx→4 g(x) with the condition x 6= 4 replaced by x < 4.
Similarly for limx→4+ g(x) the condition is x > 4.)
Since g(x) = x2 − c2 when x < 4 it follows that limx→4− g(x) = limx→4− x2 − c2 =
16 − c2 , and since g(x) = cx + 20 when x > 4 it follows that limx→4+ g(x) = g(4) =
4c + 20. Equating these gives 16 − c2 = 4c + 20, that is, c2 + 4c + 4 = 0. The only
solution is c = −2, and so we conclude that lim g(x) = g(4) if and only if c = −2.
x→4
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