Sample Calculus Problems Part 1: Single Variable Functions 1 Part 2: Multi-Variable Functions 75 Part 3: Sequences and Series 126 Part 4: Vector Analysis 151 Last revision: August 25, 2021 [This page is intentionally left blank.] Part 1: Single Variable Functions √ 4 − 3x + 1 1. Evaluate the limit lim .(Do not use L’Hôpital’s Rule.) x→5 x2 − 7x + 10 Solution: √ √ √ 4 − 3x + 1 (4 − 3x + 1)(4 + 3x + 1) lim = lim √ x→5 x2 − 7x + 10 x→5 (x2 − 7x + 10)(4 + 3x + 1) 16 − (3x + 1) = lim √ x→5 (x2 − 7x + 10)(4 + 3x + 1) 15 − 3x = lim √ x→5 (x − 5)(x − 2)(4 + 3x + 1) 3(5 − x) = lim √ x→5 (x − 5)(x − 2)(4 + 3x + 1) −3 = lim √ x→5 (x − 2)(4 + 3x + 1) − = 3√ (5 − 2)(4 + 3 ⋅ 5 + 1) − = 3 3 ⋅ 8 = −1 8 Remark: “=” is the most frequently used verb in mathematics. It was introduced in 1557 by Robert Recorde “to avoid the tedious repetition of the words ‘is equal to’.” It is important to use the equal sign correctly. • To introduce “ ”, the phantom equal sign, to avoid the tedious repetition of the symbol “=” is not a good idea. The solution above should not go like: √ √ √ 4 − 3x + 1 (4 − 3x + 1)(4 + 3x + 1) lim lim √ x→5 x2 − 7x + 10 x→5 (x2 − 7x + 10)(4 + 3x + 1) 15 − 3x lim √ x→5 (x − 5)(x − 2)(4 + 3x + 1) 3(5 − x) lim √ x→5 (x − 5)(x − 2)(4 + 3x + 1) 1 • One must also not use other symbols, which have completely different meanings, in place of “=”. The solution above should not go like: √ √ √ 4 − 3x + 1 (4 − 3x + 1)(4 + 3x + 1) lim ⇒ lim √ x→5 x2 − 7x + 10 x→5 (x2 − 7x + 10)(4 + 3x + 1) 15 − 3x → lim √ x→5 (x − 5)(x − 2)(4 + 3x + 1) 3(5 − x) lim √ x→5 (x − 5)(x − 2)(4 + 3x + 1) • The equal sign always stands between two things, although sometimes one of these things are at the end of the previous line or at the beginning of the next line. The solution above should not start like: √ √ (4 − 3x + 1)(4 + 3x + 1) = lim √ x→5 (x2 − 7x + 10)(4 + 3x + 1) √ √ (4 − 3x + 1)(4 + 3x + 1) This begs the question: What is equal to lim √ ? x→5 (x2 − 7x + 10)(4 + 3x + 1) • The equal sign can be used between two functions when we deal with identities, like x2 − 1 = x + 1 for all x =~ 1 x − 1 or when we deal with equations, like Find all x such that x2 = 4. Therefore we can not just drop some of the limit signs in the solution above to make it look like: √ √ √ 4 − 3x + 1 (4 − 3x + 1)(4 + 3x + 1) lim = √ 7 x→5 x2 − 7x + 10 (x2 − 7x + 10)(4 + 3x + 1) ⋮ − = 3√ (x − 2)(4 + 3x + 1) −3 = √ 7 (5 − 2)(4 + 3 ⋅ 5 + 1) − = 3 3 ⋅ 8 = −1 8 −3 The equalities on the lines marked with 7 are not correct. √ is (x − 2)(4 + 3x + 1) 1 −3 not equal to − because, for instance, if we let x = 1 then √ = 8 (x − 2)(4 + 3x + 1) −3 1 1 √ = =~ − . (1 − 2)(4 + 3 ⋅ 1 + 1) 2 8 2 x2 2. Let m be the slope of the tangent line to the graph of y = at the point (−3; −9). x + 2 Express m as a limit. (Do not compute m.) Solution: The slope m of the tangent line to the graph of y = f(x) at the point (x0; f(x0)) is given by the limit f(x) − f(x ) m = lim 0 → x x0 x − x0 or equivalently by the limit f(x + h) − f(x ) m = lim 0 0 : h→0 h Therefore two possible answers are 2 (− + )2 x − (− ) 3 h − (− ) + 9 − + + 9 m = lim x 2 = lim 3 h 2 : x→−3 x − (−3) h→0 h f(x) 3. Suppose that lim f(x)= ~ 0 and lim g(x) = 0. Show that lim does not exist. x→c x→c x→c g(x) f(x) f(x) Solution: Assume that lim exists, and let L = lim . Then by the product x→c g(x) x→c g(x) rule for limits we obtain f(x) f(x) lim f(x) = lim ⋅ g(x) = lim ⋅ lim g(x) = L ⋅ 0 = 0 : x→c x→c g(x) x→c g(x) x→c This contradicts the fact that lim f(x)= ~ 0. Therefore our assumption cannot be x→c f(x) true: lim does not exist. x→c g(x) 4. Suppose that lim f(x) = 0 and there exists a constant K such that Sg(x)S ≤ K for all x =~ c in x→c some open interval containing c. Show that lim (f(x)g(x)) = 0. x→c Solution: We have Sf(x)g(x)S = Sf(x)S ⋅ Sg(x)S ≤ Sf(x)S ⋅ K in some open interval around c. Therefore −KSf(x)S ≤ f(x)g(x) ≤ KSf(x)S : Now applying the Sandwich Theorem and using the fact that lim Sf(x)S = 0 we obtain x→c the result. 3 5. Determine the following limits if lim f(x) = A and lim f(x) = B. x→0+ x→0− a. lim f(x2 − x) b. lim f(x2) − f(x) c. lim f(x3 − x) x→0− x→0− x→0+ d. lim f(x3) − f(x) e. lim f(x2 − x) x→0− x→1− Solution: a. If x < 0, then x2 > 0 and −x > 0. Therefore x2 − x > 0 for x < 0, and x2−x approaches 0 from the right as x approaches 0 from the left. lim f(x2 − x) = A. x→0− b. Since x2 > 0 for x < 0, x2 approaches 0 from the right as x approaches 0 from the left. Hence lim f(x2) − f(x) = lim f(x2) − lim f(x) = A − B. x→0− x→0− x→0− c. For 0 < x < 1, we have x3 < x and x3 − x < 0. So x3 − x approaches 0 from the left as x approaches 0 from the right. Therefore lim f(x3 − x) = B. x→0+ d. Since x3 < 0 for x < 0, x3 approaches 0 from the left as x approaches 0 from the left. Hence lim f(x3) − f(x) = lim f(x3) − lim f(x) = B − B = 0. x→0− x→0− x→0− e. For 0 < x < 1 we have x2 < x and x2 − x < 0. x2 − x approaches 0 from the left as x approaches 1 from the left. Hence lim f(x2 − x) = B. x→1− 6. Let Q be the point of intersection in the first quadrant of the circle C1 with equation 2 2 2 2 2 (x − 1) + y = 1 and the circle C2 with equation x + y = r . Let R be the point where the line passing through the points P (0; r) and Q intersects the x-axis. Determine what happens to R as r → 0+. 4 Solution: Subtracting x2 + y2 = r2 from (x − 1)2 + y2»= 1 we obtain x = r2~2, and substituting this back in x2 + y2 = r2 gives us Q(r2~2; r2 − r4~4). Let R(a; 0) be the coordinates of R and let S be the foot of the perpendicular from Q to the x-axis. Since the triangles RSQ and ROP are similar we have − 2~ »a r 2 = a r2 − r4~4 r and hence r3~2 a = » : r − r2 − r4~4 Then r3~2 lim a = lim » r→0+ r→0+ r − r2 − r4~4 r3~2 » = lim ⋅ (r + r2 − r4~4) r→0+ r2 − (r2 − r4~4) » = 2 lim (1 + 1 − r2~4) r→0+ » = 2 ⋅ (1 + 1 − 02~4) = 4 : Therefore R approaches the point (4; 0) as r → 0+. 1 7. Use the formal definition of the limit to show that lim = 2. x→1~2 x Solution: Given " > 0 we want to find δ > 0 such that 1 1 0 < Vx − V < δ Ô⇒ V − 2V < ": (⋆) 2 x We will do this in two different ways. 1 “Solve the Inequality” Method: First we solve V − 2V < " for x. x 1 1 V − 2V < " ⇐⇒ 2 − " < < 2 + " x x The next step depends on whether 2 − " is positive, zero or negative. If 2 − " > 0, that is if " < 2, then 1 1 1 2 − " < < 2 + " ⇐⇒ > x > : x 2 − " 2 + " 5 If 2 − " = 0, that is if " = 2, then 1 1 2 − " < < 2 + " ⇐⇒ x > : x 2 + " If 2 − " < 0, that is if " > 2, then 1 1 1 2 − " < < 2 + " ⇐⇒ x > or − > x : x 2 + " " − 2 6 1 Next we choose δ in such a way that every x satisfying the condition 0 < Vx − V < δ 2 1 lies in the solution set of V − 2V < ", and therefore the implication in (⋆) holds. In x all three cases choosing a delta such that 0 < δ ≤ 1~2 − 1~(2 + ") = "~(4 + 2") achieves this. The Estimation Method: Suppose 0 < Sx − 1~2S < δ for some δ > 0. Then 1 2Sx − 1~2S 2δ V − 2V = < : x SxS SxS At this point let us also decide to choose δ to satisfy δ ≤ 1~4. (Why 1~4?) Then 0 < Sx − 1~2S < δ Ô⇒ 1~2 − δ < x < 1~2 + δ Ô⇒ 1~4 < x < 3~4 Ô⇒ 4 > 1~x > 4~3 Ô⇒ 1~SxS < 4 and therefore 1 2δ V − 2V < < 8δ : x SxS Hence for a given " > 0 if we choose δ to satisfy δ ≤ "~8 (as well as δ ≤ 1~4) then 1 " we will have V − 2V < 8δ ≤ 8 ⋅ = " and (∗) will hold.