November 3, 2005
Lecturer: Dr Martin Kurth
Michaelmas Term 2005
Problem Sheet 4 – Solutions
1.
lim f (x)
x→0−
=
lim (−x3 + 5)
x→0−
3
= −0 + 5
= 5
lim f (x)
x→0+
=
lim (−x3 + 5)
x→0+
3
= −0 + 5
= 5
lim f (x)
x→2−
=
lim (−x3 + 5)
x→2−
3
= −2 + 5
= −3
lim f (x) =
x→2+
lim (2x2 − 11)
x→2+
2
= 2 · 2 − 11
= −3
As the left hand and right hand limits agree at both points, the limits
exist both at x = 0 and at x = 2
2. Let’s start by giving our function a name (say f ):
cos x 2
f (x) =
+ 1.
x
So far, we know the limits of k and 1/x, and we know that the rules for
sums and products of limits apply, so we have to find two functions g and
h with
g(x) ≤ f (x) ≤ h(x)
composed of those simple functions to apply the Sandwich Theorem. As
cos x 2
1≤
+ 1,
x
g(x) = 1
is a good candidate. To find a suitable upper bound h(x), we notice that
| cos x| ≤ 1,
1
and thus
cos2 x ≤ 1.
This means that
cos x 2
1
+ 1 ≤ 2 + 1,
x
x
and we can choose this as our h(x). Now we can calculate the limits of g
and h:
lim g(x) = lim 1 = 1,
x→∞
x→∞
1
lim h(x) = lim
+ 1 = 1.
x→∞
x→∞ x2
Thus we have
lim g(x) = lim h(x) = 1,
x→∞
x→∞
and by the Sandwich Theorem
lim f (x) = 1.
x→∞
3. The first thing to notice here is that
lim
θ→π/4+
cos2 (θ − π/4) = lim cos2 (θ),
θ→0+
and we introduce the notation
f (θ) = cos2 (θ).
Again, we only know limits of k and θ, and have to find functions g and
h with
g(θ) ≤ f (θ) ≤ h(θ),
that are combinations of such simple functions. As we are looking at the
right-hand limit at 0, we can restrict ourselves to
0<θ<
π
,
2
and study the situation in this picture:
r
y
θ
x
2
s
Now we have
cos2 (θ) =
x2
r2
and
x2 = r 2 − y 2
(from Pythagoras’ Theorem), such that
cos2 (θ) = 1 −
y2
.
r2
Furthermore, we know from our picture that
y≤s
and thus
y 2 ≤ s2 ,
which gives us
cos2 (θ) ≥ 1 −
As the angle θ is defined as
s2
.
r2
s
θ= ,
r
we have
cos2 (θ) ≥ 1 − θ2 .
This gives us a candidate for our function g:
g(θ) = 1 − θ 2 .
As we know that
cos2 (θ) ≤ 1,
we choose
h(θ) = 1.
Now we have
g(θ) ≤ f (θ) ≤ h(θ),
and
lim g(θ) = lim (1 − θ2 ) = 1,
θ→0+
θ→0+
lim h(θ) = lim 1 = 1,
θ→0+
θ→0+
and thus
lim f (θ) = 1.
θ→0+
3