false

Class Notes
(839,092)

Canada
(511,185)

University of Toronto Scarborough
(31,651)

Mathematics
(870)

MATA31H3
(103)

Leo Goldmakher
(28)

Lecture

Department

Mathematics

Course Code

MATA31H3

Professor

Leo Goldmakher

Description

U NIVERSITY OF TORONTO SCARBOROUGH
MATA31H3 : Calculus for Mathematical Sciences I
FINAL EXAMINATION
December 20, 2012
Duration – 3 hours
Aids: none
NAME (PRINT):
Last/Surname First/Given Name (and nickname)
STUDENT NO: KEY
TUTORIAL:
Tutorial section # Name of TA
Qn. # Value Score
COVER PAGE 5
1 25
2 10
3 15
4 10
5 10
6 25
Total 100
TOTAL:
Please read the following statement and sign below:
I understand that any breach of academic integrity is a violation of The Code of Behaviour on Academic Matters. By
signing below, I pledge to abide by the Code.
SIGNATURE: Final Exam MATA31H3 page 2 of 10
(1) (a) (15 points) Prove that lim x ▯3x+3 = 1. For this part of the problem you may not use
x!2
any theorems from lecture.
Fix ▯ > 0, and suppose x satisﬁes
n o
(*) 0 < jx ▯ 2j < min 1; ▯ :
2
It follows that jx ▯ 2j < 1, i.e. that ▯1 < x ▯ 2 < 1. Thus, for all x satisfying
(*), we have 0 < x ▯ 1 < 2. On the other hand, (*) implies that jx ▯ 2j < , so
2
2 ▯
j(x ▯ 3x + 3) ▯ 1j = jx ▯ 1j ▯ jx ▯ 2j < 2 2 = ▯:
This proves the claim. ▯
(b) (10 points) Suppose f : R ▯! R satisﬁes x ▯ 1 ▯ f(x) ▯ x ▯ 3x + 3 for all x 2 R.
Determine (with proof) the value of lx!2f(x). For this part of the problem you may use
any theorems from lecture.
First, note that lim x ▯ 1 = 1. [For any ▯ > 0, we have 0 < jx ▯ 2j < ▯ =)
x!2
j(x▯1)▯1j = jx▯2j < ▯.] Next, from part (a) we know that lim x ▯3x+3 = 1.
x!2
The squeeze theorem therefore implies that lim f(x) exists and is equal to 1.
x!2
continued on page 3 Final Exam MATA31H3 page 3 of 10
Page intentionally left blank for scratch work.
continued on page 4 Final Exam MATA31H3 page 4 of 10
(2) Consider the function
g : R ▯! [0;1)
x2
x 7▯! 2
1 + x
(a) (5 points) Is g injective? If so, prove it; if not, provide a counterexample.
1
No, g is not injective. For example, g(12 =g(▯1).
(b) (5 points) Is g surjective? If so, prove it; if not, provide a counterexample.
Yes, g is surjective. In other words, for every y 2 [0;1), there exists some
real number which gets mqpped to y by the function g. To see this, pick an
y
arbitrary y 2 [0;1). Then 1▯y2 R, since y ▯ 0 and 1 ▯ y > 0. Moreover,
▯ ▯
r y y y
g = 1▯yy = 1▯y = y:
1 ▯ y 1 + 1▯y 1▯y
continued on page 5 Final Exam MATA31H3 page 5 of 10
(3) (15 points) Let
(
h(x) := sinx if x 2 Q
3 if x 62 Q:
Prove that lim h(x) does not exist.
x!2
We proceed by contradiction. Suppose that the limit does exist, say,
lim h(x) = L:
x!2
It follows that there exists some constant ▯ > 0 such that
1
(y) jh(x) ▯ Lj < 2 8x 2 (2;2 + ▯):
From the density theorem, we know that there exists a rational number a 2 (2;2+▯)
and an irrational number b 2 (2;2 + ▯). The inequality (y) thus implies that
jh(a) ▯ Lj < 1 and jh(b) ▯ Lj < 1
2 2
so that triangle inequality implies
(z) jsina ▯ 3j = j(sina ▯ L) + (L ▯ 3)j ▯ jsina ▯ Lj + jL ▯ 3j < 1:
On the other hand, since sina ▯ 1, we have
jsina ▯ 3j = 3 ▯ sina ▯ 2;
which contradicts (z). ▯
continued on page 6 Final Exam MATA31H3

More
Less
Unlock Document

Related notes for MATA31H3

Only pages 1,2 and half of page 3 are available for preview. Some parts have been intentionally blurred.

Unlock DocumentJoin OneClass

Access over 10 million pages of study

documents for 1.3 million courses.

Sign up

Join to view

Continue

Continue
OR

By registering, I agree to the
Terms
and
Privacy Policies

Already have an account?
Log in

Just a few more details

So we can recommend you notes for your school.

Reset Password

Please enter below the email address you registered with and we will send you a link to reset your password.