MA552 Analysis Class Test 2
Name:
Attempt all FOUR questions.
. Evaluate the following limits, justifying your answers.
Q1
(i)
tan(2x)
x→0 tan(x)
lim
(ii)
1
lim
n→∞
3n − 1
1
7n − 1
[Recall that ax := ex log a .]
[5 marks]
(i) Let f (x) = tan(2x) and g(x) = tan(x). Both f and g are dierentiable and
f (0) = g(0) = 0. So we may apply L'Hôpital's Rule to say:
tan(2x)
2 sec2 (2x)
= lim
x→0 tan(x)
x→0 sec2 (x)
lim
provided the right hand side limit exists. It does by the continuity of sec at 0
and equals:
2 sec2 (0)
2
= = 2.
2
sec (0)
1
(ii) Write x = 1/n. Then x → 0+ as n → ∞. Therefore it is enough to calculate:
3x − 1
x→0 7x − 1
Let f (x) = 3x − 1 and g(x) = 7x − 1. Both f and g are dierentiable and
f (0) = g(0) = 0. So we may apply L'Hôpital's Rule to say:
lim
3x − 1
log(3)3x
=
lim
x→0 log(7)7x
x→0 7x − 1
lim
provided the right hand side limit exists. (Here we use that the derivative of ax
is log(a)ax .) It does exist by continuity of 3x and 7x and equals:
log(3)
log(3)30
=
.
0
log(7)7
log(7)
[Note: In (ii) you may also use the version of L'Hôpital's Rule that deals with
1
limits as n → ∞. If you did this then note that if F (x) = 3 x − 1 then F 0 (x) =
log(3)(− x12 )31/x .]
1
. Suppose that f : [a, b] → R is a continuous function that is
dierentiable on (a, b) which satises f 0 (x) = 3 for all x ∈ (a, b). Show
that f (x) = 3x − 3a + f (a) for all x ∈ (a, b).
Q2
[5 marks]
We apply the Mean Value Theorem to f on the interval [a, x]. It is valid
because f is continuous on [a, b] and dierentiable on (a, b). We obtain that:
There exists some point c ∈ (a, x) such that
f 0 (c) =
f (x) − f (a)
.
x−a
Now we know that f 0 (c) = 3 so this gives that 3 =
f (x) − f (a), f (x) = 3x − 3a + f (a).
f (x)−f (a)
,
x−a
that is 3(x − a) =
[Note:You could also use Taylor's Theorem on this question. This gives that:
f (x) = f (a) +
(x − a)f 0 (c)
1!
for some c ∈ (a, x). But note that this is just a rearrangement of the conclusion
of the Mean Value Theorem! In the question you are only told that f is dierentiable on (a, b) and therefore you don't know if f 0 (a) exists, so you should not
use Taylor's Theorem with any larger value of n. ]
2
. Let f (x) = 2|x| + 1 and P denote the partition
Q3
1
1
−1 < − < 0 < < 1
3
2
of [−1, 1]. Calculate the lower Riemann sum L(f, P) and the upper
Riemann sum U (f, P).
[4 marks]
L(f, P) =
4
X
mi (ξi − ξi−1 )and U (f, P) =
i=1
4
X
Mi (ξi − ξi−1 )
i=1
where ξ0 = −1, ξ1 = − 31 , ξ2 = 0 ξ3 =
1
2
and ξ4 = 1. And:
mi = inf{sin(x) : x ∈ [ξi−1 , ξi ]}
Mi = sup{sin(x) : x ∈ [ξi−1 , ξi ]}.
So here m1 = 35 , m2 = 1 = m3 , m4 = 2 and M1 = 3, M2 = 53 , M3 = 2, M4 = 3.
So we obtain:
L(f, P) =
and
52
1
1
1
53
+1 +1 +2 =
33
3
2
2
18
2 51
1
1
91
U (f, P) = 3 +
+2 +3 = .
3 33
2
2
18
3
. For each of the following statements, decide if it is true or false, giving
reasons for your answers.
Q4
(i) Suppose f : [1, 2] → R is dened by f (x) = ex sin(x) + x2 cos(x) − 3,
then there exist c, d ∈ [1, 2] such that
f (c) ≤ f (x) ≤ f (d) for all x ∈ [1, 2].
(ii) If g : [a, b] → [c, d] is a continuous function which is both
dierentiable on (a, b) and invertible. Then g 0 (x) 6= 0 for all x ∈ [a, b].
[6 marks]
(i) This is true.
Using that sin x, cos x, ex , x and x2 are all continuous functions, and the fact
that any composition, product or sum of continuous functions is continuous, we
obtain that f is continuous. And a continuous function on a closed bounded
interval such as [1, 2] is bounded and attains its bounds by the Extreme Values
Theorem, giving us the result.
(ii) This is false.
Here are some counterexamples:
1. Consider f (x) = x3 dened on [−1, 1]. This is dierentiable and has
inverse x1/3 but f 0 (x) = 3x2 is zero at x = 0.
2. Or consider f (x) = cos x on [0, π]. This is dierentiable and invertible,
but f 0 (x) = − sin x is zero at x = 0 and at x = π .
[Note that the Inverse Function Theorem requires that a function has non-zero
derivative in order to construct a dierentiable inverse. In the examples above,
the inverse is not dierentiable at some points.]
4