MATH 1A MIDTERM 1 (8 AM VERSION) SOLUTION. (Last edited October 18, 2013 at 5:06pm.) lim

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1 MATH A MIDTERM (8 AM VERSION) SOLUTION (Last edited October 8, 03 at 5:06pm.) Problem. (i) State the Squeeze Theorem. (ii) Prove the Squeeze Theorem. (iii) Using a carefully justified application of the Squeeze Theorem, find ( ) 7 sin. (i) If f, g, h are real-valued functions such that f() g() h() () when is near a (ecept possibly at a) and if a f() a h() = L, then a g() = L. The precise meaning of near a (ecept possibly at a) is: there eists some δ > 0 such that () holds for all satisfying 0 < a < δ. (Remark: f, g, h do not need to be continuous.) (ii) Suppose f, g, h satisfy () when 0 < a < δ for some fied δ > 0. Also suppose that a f() a h() = L. Let ε > 0. Since a f() = L, there eists δ > 0 such that if 0 < a < δ, then f() L < ε. Also, since a h() = L, there eists δ 3 > 0 such that if 0 < a < δ 3, then h() L < ε. Put δ = min{δ, δ, δ 3 }. Suppose 0 < a < δ. Then (), L ε < f() < L + ε and L ε < h() < L + ε hold. Hence g() h() < L + ε. Also L ε < f() g(). Thus L ε < g() < L + ε, so g() L < ε. Hence a g() = L. (iii) Set f() = 7, g() = 7 sin, and h() = 7. (Note that g is only defined for 0.) Note that sin () for all 0. Multiplying both sides of () by 7, we have 7 sin = 7 sin 7 which implies 7 7 sin 7 which is just (). (Here we can take δ to be any positive real number.) Note that f() = h() = 0. Hence, by the Squeeze Theorem, g() = 0. (Remark: it was commonly claimed that 7 7 sin 7

2 MATH A MIDTERM (8 AM VERSION) SOLUTION for all 0, but it isn t true if < 0; we have instead 7 7 sin 7 so the Squeeze Theorem does not apply.) Problem. Evaluate +. You should show your reasoning carefully, however you may use any of the it laws without eplanation or proof. We have + ( )( + ) ( + ) ( + ) ( ) ( + + ) ( + + ) + + = = + =. Problem 3. (i) Let f be a real valued function, and let a and L be real numbers. What does it mean to say that a f() = L? (ii) Prove carefully, using the definition you gave in part (i), that ( ) =. (i) The statement a f() = L means for every ɛ > 0, there eists δ > 0 such that if 0 < a < δ then f() L < ɛ. (ii) Let ɛ > 0. Put δ = min{ ɛ 000, ɛ 000 }. Suppose 0 < < δ. Then 0 < < 0 < < 000. Thus we have = ( ) + 4( ) + 6 = (by the triangle inequality) = ( 000 ) + 4( 000 ) + 6 (since < 000 ) < and

3 MATH A MIDTERM (8 AM VERSION) SOLUTION 3 So ( ) ( ) = = < ɛ < ɛ. Thus ( ) =. Problem 4. (i) State carefully the Intermediate Value Theorem. (ii) Prove that there is a root of the equation in the interval (, ). 3 = 3 (i) Let f be a continuous function on the interval [a, b]. Suppose f(a) f(b) and let N be a value strictly between f(a) and f(b). Then there eists some c (a, b) such that f(c) = N. (Remark: f needs to be continuous on the closed interval [a, b], not just on the open interval (a, b), and we can guarantee that such c eists in the open interval (a, b), not just the closed interval [a, b].) (ii) Let f() = 3 3. Since f is the sum of a polynomial and an eponential function, it is continuous on the interval [, ] (it is even continuous on the entire real line, but we only need it to be continuous on [, ]). Observe that f() = () 3 3 = f() = () 3 3 = 7. Set N = 0; it is strictly between f() and f(). By the Intermediate Value Theorem, there eists c (, ) such that f(c) = 0. This implies that c is a solution to 3 = 3. Problem 5. The figure below shows the graph of y = f() when 5 + if < π tan if π < 0 but π 0 if = π if 0 < f() = 3 if < 4 if = 4 + if 4 < < 7 but 5 if = 5 ln if 7 For each of the following statements, indicate if it is true or false. (i) 4 f() = 3

4 4 MATH A MIDTERM (8 AM VERSION) SOLUTION (ii) 5 + f() = (iii) 5 f() = (iv) 3 f() eists (v) π + f() = (vi) f() = (vii) The graph y = f() has a horizontal asymptote at y = 0. (viii) The graph y = f() has two horizontal asymptotes. (i) The graph y = f() has two vertical asymptotes. () f() is continuous at = 0. (i) f() is continuous at =. (ii) f() is continuous on the interval [, ]. Proof. (i) True. We have 4 f() 4 3 = 3 and 4 + f() 4 +( + ) = (4 5) + = 3. This implies 4 f() = 3. (Notice that the it is still 3 even if f(4) 3.) (ii) True. We prove that for every N > 0 there eists δ > 0 such that if 0 < 5 < δ then f() > N, which is what the statement = means. Let N > 0. Set δ = min{ N, } if N > and δ = otherwise. (Having δ is nice because it ensures that f() = + for all satisfying 0 < 5 < δ). Suppose 0 < 5 < δ. If N >, then 0 < 5 < N, so N <, thus N < + = f(). If N, then 0 < 5 <, so ; thus N < 3 + = f(). (iii) True. We prove that for every N > 0 there eists δ > 0 such that if 0 < 5 < δ then f() > N, which is what the statement 5 + = means. Let N > 0. Set δ = min{ N, } if N > and δ = otherwise. Suppose 0 < 5 < δ. If N >, then 0 < 5 < N, so N <, thus N < + = f(). If N, then 0 < 5 <, so ; thus N < 3 + = f(). (iv) True. Since f() = tan on the interval ( π, π ), which contains 3, f is continuous at = 3 and f( 3) 3 f() (by the definition of continuity). (v) False. We actually have π + f() = and π f() =. (vi) True. Note that f() = ln on [7, ); for any positive N > 0, we have ln > N for all > e N. (vii) True. We interpret the statement the graph y = f() has a horizontal asymptote at y = 0 to mean at least one of f() = 0 and f() = 0 holds. In this case we have f() = 0. (viii) False. The graph y = f() has eactly one horizontal asymptote, namely at y = 0. Although it might seem like there is a horizontal asymptote at y 3.4, it is not the case since f() =. The graph shows only a finite section of ln, which grows increasingly slowly with respect to. (i) True. The two vertical asymptotes are at = π and = 5. () True. We have f() tan = 0 and + f() + = 0, so f() is continuous at = 0. (i) False. We have f() = and + f() + 3 = 3, and these two one-sided its do not agree, so f() is not continuous at =. (ii) True. Professor Coward has addressed this issue; his definition of continuity of f on an interval [a, b] is: f is continuous in the usual sense for each point (a, b), and f(a) a + f() and f(b) b f(). This is a weaker condition than the alternative definition, which requires additionally that a f() eists and is equal to f(a) and b f() eists and is equal to

5 MATH A MIDTERM (8 AM VERSION) SOLUTION 5 f(b). Importantly, the Intermediate Value Theorem is still true for functions which satisfy the weaker condition. Problem 6. Using the it definition of derivative, show that if f() =, then f () =. Set f() =. We have f( + h) f() ( + h) h 0 h h 0 h h 0 + h + h h h 0 + h =. Thus f () =. Problem 7. What is log 4 8? Let = log 4 8. Recall that the epression log 4 8 is defined to be the unique number such that 4 log 4 8 = 8. Thus 4 = 8. Write 4 = and 8 = 3 so that = 3. Thus = 3, so = 3.

Calculus I (part 1): Limits and Continuity (by Evan Dummit, 2016, v. 2.01)

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