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These are the Notes of Exam Feedback of Real Analysis which includes Value of Derivative, Triangle Inequality, Sandwich Rule, Product Rule for Limits etc. Key important points are: Triangle Inequality, Sandwich Rule, Intermediate Value Theorem, Standard Results, Rational Functions, Continuous Function, Rolle’s Theorem, Uniqueness of Solution
Typology: Exams
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(a) By verifying the appropriate definitions prove
(i) (^) xlim→ 2 (2x^2 − x + 1) = 7,
(ii) (^) xlim→∞
x^2 + x + 1
(b) Suppose that f, g and h are three functions such that
h (x) ≤ f (x) ≤ g (x)
for all x in some deleted neighbourhood of a ∈ R.
Prove that if limx→a h (x) = limx→a g (x) = L then limx→a f (x) = L. (c) Calculate
lim x→ 0
x^2 sin
x
2 − cos x
2 x^2 − x − 6
= |(2x + 3) (x − 2)| < 9 |x − 2 | ≤ 9 (ε/9) = ε.
Hence we have verified the definition that limx→ 2 (2x^2 − x + 1) = 7. Similar to problem on problem sheet [5 marks] (ii) Let ε > 0 be given. Choose X = 1/ε > 0. Assume x > X. Then 1 x^2 + x + 1
x
on “throwing away” x^2 + 1 > 0,
= ε.
Hence we have verified the definition that lim x→∞ 1 / (x^2 + x + 1) = 0.
Similar to problem on problem sheet [5 marks] (b) There exists δ 0 > 0 such that if 0 < |x − a| < δ 0 then h (x) ≤ f (x) ≤ g (x).
Let ε > 0 be given. The definition of limx→a h (x) = L implies there exists δ 1 > 0 such that if 0 < |x − a| < δ 1 then |h (x) − L| < ε/3. The definition of limx→a g (x) = L implies there exists δ 2 > 0 such that if 0 < |x − a| < δ 2 then |g (x) − L| < ε/3. Let δ = min (δ 0 , δ 1 , δ 2 ). Assume x satisfies 0 < |x − a| < δ. For such x all h (x) ≤ f (x) ≤ g (x), |h (x) − L| < ε/3 and |g (x) − L| < ε/3 hold.
Consider |f (x) − L| = |f (x) − h (x) + h (x) − L| ≤ |f (x) − h (x)| + |h (x) − L|
by the triangle inequality. The assumption h (x) ≤ f (x) ≤ g (x) implies 0 ≤ f (x) − h (x) ≤ g (x) − h (x). Since the differences are both positive they equal their moduli, and so |f (x) − h (x)| ≤ |g (x) − h (x)|. Thus
|f (x) − h (x)| + |h (x) − L| ≤ |g (x) − h (x)| + |h (x) − L| = |g (x) − L + L − h (x)| + |h (x) − L| ≤ |g (x) − L| + |L − h (x)| + |h (x) − L| by the triangle inequality again, < ε/3 + ε/3 + ε/3 = ε.
(a) (i) If limx→a g (x) = L exists and f is continuous at L prove that
xlim→a f^ (g^ (x)) =^ f
x^ lim→a g^ (x)
= f (L).
(ii) Show that cos
x − 2 x^2 − 4 x + 8
is continuous on R.
All standard results on continuity you assume should be carefully stated. (b) Carefully state the Intermediate Value Theorem. Prove that x sin x
cos x
= π
has a solution with x ∈ (0, π/2).
Prove that the solution found is unique within (0, π/2). State carefully any results you use to prove this. It may be of use to assume that tan x > x in (0, π/2).
implies there exists δ 1 such that
if |y − L| < δ 1 then |f (y) − f (L)| < ε. (1)
Apply the definition of limx→a g (x) = L with ε replaced by δ 1 to find δ > 0 such that
if 0 < |x − a| < δ then |g (x) − L| < δ 1. (2)
Combine (2) and (1) with y = g (x) to get
if 0 < |x − a| < δ then |f (g (x)) − f (L)| < ε.
Thus we have verified the definition that limx→a f (g (x)) = f (L). Bookwork [6 marks]
(ii) We assume the results that rational functions are continuous at all points at which they are defined while cos is continuous on R.
We need check that (x − 2) / (x^2 − 4 x + 8) is defined on R. Either note that x^2 − 4 x + 8 = (x − 2)^2 + 4 > 0 and so never zero, or note that because (−4)^2 − 4 × 8 < 0 the roots are complex and so, again, x^2 − 4 x + 8 is never zero for x ∈ R.
Apply part (i). [3 marks]
(b) Intermediate Value Theorem. Suppose that f is a function continuous on a closed interval [a, b] and that f (a) 6 = f (b). If γ is some number between f (a) and f (b) then there must be at least one c : a < c < b for which f (c) = γ. Bookwork [2 marks]
Let f (x) = x/ sin x + 1/ cos x. To use the Intermediate Value Theorem we need a closed interval, which cannot be [0, π/2] since f is not defined at the end points. Instead we look at points within this interval. We find that
f
(π 6
2 π 6
≈ 2. 20 < π,
f
(π 3
2 π 3
Hence, by applying the Intermediate Value Theorem with γ = π on [π/ 6 , π/3] , we find there is a solution to f (x) = π somewhere in (0, π/2). Similar to problem covered in lectures [5 marks]
(a) State the Mean Value Theorem. Prove
(i) sin x < x for x > 0 and (ii) ln (1 + x) < x for x > 0.
Deduce e−x^ sin x <
x 1 + x
for x > 0.
(b) Let f, g : A → R both be differentiable at a ∈ A. Prove, using the rules of limits, that
(f g)′^ (a) = f ′^ (a) g (a) + f (a) g′^ (a).
(c) Calculate the Taylor polynomial
T 7 , 0
e−x^ sin x
Prove, using Lagrange’s form for the error, that ∣∣ e−x^ sin x − T 7 , 0
e−x^ sin x
for |x| ≤ 10 −^1.
Hint, You should use the result from Part a in bounding the result that arises from the application of Lagrange’s form of the error.
continuous on [a, b] and differentiable on (a, b). Then there exists c ∈ (a, b) such that f (b) − f (a) b − a
= f ′^ (c).
Bookwork [2 marks] (i) With f (x) = sin x we get, for some 0 < c < x,
sin x − sin 0 x − 0
= cos c ≤ 1.
So for x > 0 we can multiply up to get result. From Lectures [1 mark]
(ii) With f (x) = ln (1 + x) we get, for some 0 < c < x,
ln (1 + x) − 0 x − 0
1 + c
So for x > 0 we can multiply up to get result. From Lectures [1 mark]
Exponentiating ln (1 + x) < x we get 1 + x < ex^ or e−x^ < 1 / (1 + x). Combine with (i) to get
e−x^ sin x <
x 1 + x
for x > 0. [1 mark]
(b) Consider
f (x) g (x) − f (a) g (a) x − a
f (x) g (x) − f (a) g (x) + f (a) g (x) − f (a) g (a) x − a
= g (x)
f (x) − f (a) x − a
g (x) − g (a) x − a
Then by the sum and product rules for limits, allowable since the final limits exist, and the fact that a function is continuous when differentiable,
(f g)′^ (a) = (^) xlim→a g (x) lim x→a
f (x) − f (a) x − a
g (x) − g (a) x − a = g (a) f ′^ (a) + f (a) g′^ (a)
as required. Bookwork [4 marks]
(a) Calculate the lower and upper sums
L (Qn, f ) and U (Qn, f ) ,
where f : [2, 4] , x → 2 x^2 , and
Qn =
2 ηi^ : 0 ≤ i ≤ n
is the n-th geometric partition of [2, 4], so η = 2^1 /n.
(b) Let f : [a, b] → R be a bounded function. Define the lower and upper integrals, ∫ (^) b
a
f and
∫ (^) b
a
f.
(c) Assuming that L (P, f ) ≤ U (D, f ) for all partitions P and D of [a, b], prove ∫ (^) b
a
f ≤
∫ b
a
f.
Define what is meant by saying that f : [a, b] → R, a bounded function, is Riemann Integrable.
(d) Prove that f : [2, 4] , x → 2 x^2 is Riemann integrable and find the value of the integral.
so on each subinterval [xi− 1 , xi] = [2ηi−^1 , 2 ηi] we have mi = 2
2 η(i−1)
and
Mi = 2 (2ηi)^2. Then
U (Qn, f ) =
∑^ n
i=
Mi (xi − xi− 1 )
∑^ n
i=
2 ηi
2 ηi^ − 2 ηi−^1
1 − η−^1
) ∑n
i=
η^3 i^ using ηi^ − ηi−^1 = ηi^
1 − η−^1
1 − η−^1
) (^) η^3 (η^3 n^ − 1) η^3 − 1
on summing the geometric series,
= 16 (η − 1)
η^2 (2^3 − 1) (η − 1) (η^2 + η + 1)
since ηn^ = 2
112 η^2 η^2 + η + 1
We don’t wish to do the same work twice so we relate the Lower sum to the Upper one by
L (Qn, f ) =
∑^ n
i=
mi (xi − xi− 1 )
∑^ n
i=
2 ηi−^1
2 ηi^ − 2 ηi−^1
η^2
U (Qn, f ) =
η^2 + η + 1
Similar to problem covered in lectures [8 marks]
(b) For f bounded on [a, b], the Upper Integral is ∫ b
a
f = glb {U (P, f ) : P a partition of [a, b]}
and the Lower Integral is ∫ (^) b
a
f = lub {L (P, f ) : P a partition of [a, b]}.
