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Lecture

FinSoln.pdf

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Department
Mathematics
Course Code
MATA31H3
Professor
Leo Goldmakher

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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
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