mt20101 - exam paper - 2007
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Math20101 - Exam Paper - 2007
Michael Bushell [email protected]
January 4, 2012
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1 REAL ANALYSIS 2
1 Real Analysis
Example 1.0.1 (A1). 1. By verifying the appropriate definitions
(a)limx→2
(2x2 − x + 1) = 7
Proof. Using the definition
limx→a
f (x) = L ⇔ ∀ > 0, ∃δ > 0 : 0 < |x − a| < δ ⇒ |f (x) − L| <
Let > 0 be given and choose δ = min(1, /9), then assuming
0 < |x − 2| < δ we have
|x − 2| < δ =⇒ |x − 2| < 1 as δ ≤ 1
=⇒ −1 < x − 2 < 1
=⇒ 5 < 2x + 3 < 9
=⇒ |2x + 3| < 9
it follows that
|(2x2 − x + 1) − 7| = |2x2 − x − 6|
= |2x + 3||x − 2|< 9 × /9
=
and hence we have verified the definition.
(b)
limx→∞
1
x2 + x + 1= 0
Proof. Using the definition
limx→∞
f (x) = L ⇔ ∀ > 0, ∃K > 0 : x > K ⇒ |f (x) − L| <
Let > 0 be given and choose K = 1/ > 0, then assuming x > K we have 1
x2 + x + 1− 0
≤1
x<
1
k=
Therefore, we have verified the definition, as required.
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1 REAL ANALYSIS 3
2. 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, and suppose
limx→a
h(x) = limx→a
g(x) = L
thenlimx→a
f (x) = L
Proof. Let > 0 be given, and choose δ0 > 0 such that h(x) ≤ f (x) ≤g(x) for all 0 < |x − a| < δ0, that is δ0 is the width of the deletedneighbourhood of a where this holds by hypothesis.
Choose δ1 > 0 such that |h(f ) − L| < whenever 0 < |x − a| < δ1,then L − < h(x). Similarly, choose δ2 > 0 such that |g(f ) − L| < whenver 0 < |x − a| < δ2, then g(x) < L + .
Now, let δ = min(δ0, δ1, δ2), then if 0 < |x − a| < δ, we have
L − < h(x) ≤ f (x) ≤ g(x) < L +
therefore− < f (x) − L < i.e., |f (x) − L| <
and we have verified the definition as required.
3. Evaluate
limx→0
x2 sin(1/x)
2 − cos(x)
Solution. We have −x2
≤ x2
sin(1/x) ≤ x2
, for x = 0, as | sin(1/x)| ≤1. Therefore, by the sandwich rule given above, since ±x2 → 0 asx → 0 we can conclude x2 sin(1/x) → 0 as x → 0 also. Thus
limx→0
x2 sin(1/x)
2 − cos(x)=
limx→0 x2 sin(1/x)
limx→0[2 − cos(x)]
=0
2 − 1= 0
using the limit of a quotient, and limit of a sum rules.
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1 REAL ANALYSIS 4
Example 1.0.2 (A2).
1. (a) Suppose limx→a g(x) = L and f is continuous at L, then
limx→a
f (g(x)) = f (limx→a
g(x))
Proof. . . .
(b)
cos
x − 2
x2 − 4x + 8
is continuous onR
.
Proof. The quotient of two polynomials is continuous at all pointswhere the denominator is non-zero. As x2−4x+8 = (x−2)2+4 ≥ 0for all x, it is never 0, and therefore (x − 2)/(x2 − 4x + 8) iscontinuous on R. We know that cos(x) is continuous on R, soapplying the composite rule proved above we have our desiredresult.
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2 COMPLEX ANALYSIS 5
2 Complex Analysis