Tutorial 4: SOLUTIONS
Continuity of Trigonometric Functions
1. We are asked to compute the following limits
1
(a) lim cos
. Since the cosine function is continuous, we can bring the limit
x→∞
x
inside to get:
1
cos lim
= cos(0) = 1.
x→∞ x
πx . Again since the sine function is continuous we can bring
(b) lim sin
x→∞
3 − 5x
the limit inside to get
πx π sin lim
= sin lim
= − sin(π/5) = −0.587785.
x→∞ 3 − 5x
x→∞ 3/x − 5
sin 7θ
. Multiplying above and below by 16θ/θ, we obtain
θ→0 sin 16θ
7
sin θ 16θ 7
7
7θ sin 7θ 16θ
=
lim
lim
= (1)(1) =
lim
θ→0 sin 16θ
θ→0 16θ 7θ sin 16θ
16 θ→0 7θ
16
16
(c) lim
sin h
. Multiplying above and below by 1 + cos h gives
h→0 1 − cos h
(d) lim
sin h(1 + cos h)
1 + cos h
2
sin h(1 + cos h)
= lim
= lim
= = ∞.
2
2
h→0
h→0
h→0
1 − cos h
sin h
0
sin h
lim
Hence the limit doesn’t exist.
1 − cos(5x)
. Multiplying above and below by 1 + cos(5x) gives
(e) lim
x→0 cos2 (3x) − 1
1 − cos2 (5x)
sin2 (5x)
=
lim
.
x→0 (cos2 (3x) − 1)(1 + cos(5x))
x→0 (− sin2 (3x))(1 + cos(5x))
lim
Now multiplying above and below by (5x/3x)2 gives
1
(5x)2 sin2 (5x) (3x)2
2
2
2
x→0 (3x)
(5x) sin (3x) 1 + cos(5x)
25
sin(5x) 2 3x 2 1
=−
lim
lim
lim
x→0 sin(3x)
x→0 1 + cos(5x)
9 x→0 5x
25
1
= − (1)2 (1)2
9
1 + (1)
25
=− .
18
− lim
1
(f) lim
x→0
tan(ax)
tan(bx)
(a 6= 0, b 6= 0). Writing in terms of sines and cosines gives
sin(ax) cos(bx)
ax sin(ax) bx cos(bx)
= lim
x→0 cos(ax) sin(bx)
x→0 bx
ax sin(bx) cos(ax)
a
sin(ax)
bx cos(bx) =
lim
lim
lim
x→0 sin(bx)
x→0 cos(ax)
b x→0 ax
a
= (1)(1)(1)
b
a
= .
b
2. We are asked to determine whether the piecewise function
(
sin x
x 6= 0
|x|
f (x) =
1
x=0
lim
is continuous at x = 0? For continuity at x = 0, we must have
lim f (x) = f (0) = 1.
x→0
To determine whether the two-sided limit exists, we must consider the one-sided
limits:
sin x
sin x
lim f (x) = lim
= lim
= −1
−
−
−
|x|
−x
x→0
x→0
x→0
sin x
sin x
= lim
= 1.
lim f (x) = lim
+
+
+
|x|
x
x→0
x→0
x→0
Since the one-sided limits are not equal, the two-sided limit doesn’t exist at x = 0
and so the function is not continuous there.
3. We wish to find a non-zero value of k such that
(
f (x) =
tan(kx)
x
7x +
4k 2
x<0
x≥0
is continuous at x = 0. It is clear from the definition of f (x) that
lim f (x) = f (0) = 4k 2 .
x→0+
For continuity, this must also be equal to the limit from the left, which is given by
lim f (x) = lim
x→0−
x→0−
k
sin(kx)
k
tan(kx)
= lim
=
(1) = k.
x
kx
(1)
x→0− cos(kx)
Continuity then requires
4k 2 = k,
=⇒ k(4k − 1) = 0,
=⇒ k = 0, k = 1/4.
But since we’re asked for a non-zero value of k, we have
k = 1/4.
2
4. We are asked if f (x) = x sin(1/x) is continuous at x = 0. The answer is NO since
the function is not well-defined at x = 0.
We wish to define a continuous piecewise function g(x) which is equal to f (x) for
all x 6= 0. We showed in class (by using the Squeezing Theorem) that
lim x sin(1/x) = 0
x→0
and hence the function
(
x sin(1/x)
g(x) =
0
x 6= 0
x = 0.
is continuous at x = 0 and indeed is continuous everywhere.
5. We are asked to use the Intermediate Value Theorem to show that there is at least
one root of f (x) = x − cos x in the interval [0, π/2]. Given that there is exactly
one root in this interval, we want to approximate this root to two decimal places
of accuracy.
We start by evaluating the function at the endpoints: f (0) = −1 and f (π/2) = π/2.
Since there is a sign change, according to the Intermediate Value Theorem there
is a root in the interval [0, π/2].
To approximate the root, we divide the interval into intervals of length 0.1 and
evaluate the function at each point:
x
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
f (x) −1 −0.895 −0.78 −0.655 −0.521 −0.378 −0.225 −0.0648 0.103
Since there is a sign change between 0.7 and 0.8, according to the Intermediate
Value Theorem, there is a root in the interval (0.7, 0.8). The midpoint of this
interval is an approximation of the root to one decimal place of accuracy. To
obtain two decimal places of accuracy, we repeat the algorithm one more time,
i.e., we divide (0.7, 0.8) into intervals of length 0.01 and evaluate at each partition
point. The table is as follows:
x
0.7
0.71
0.72
0.73
0.74
f (x) −0.0648 −0.04834 −0.0318 −0.015 0.00153
Since there is a sign change in the interval (0.73, 0.74), there is a root in this
interval, which we approximate, to two decimal places of accuracy, by the midpoint
of the interval, which is
x = 0.735
A plot of the function is given above.
Note: Two decimal places of accuracy means that we can only trust our answer to
2 decimal places, i.e., the first two digits of the actual root are 0.73, to determine
each remaining digit would require a further iteration of the algorithm. Of course,
the Newton-Raphson method is far more efficient than this.
3
1.5
f (x) = x − cos x
1.0
0.5
0.5
1.0
xx ∼
∼0.746128
0.739085
�0.5
�1.0
4
1.5