微甲

1. (14%, 7% each) Evaluate the following limits:

(a) lim

x→0(sin⁻¹x x )^x21 . (b) lim

x→0⁺(x^xx − 1).

Sol:

(a) let x = sin y

as x→ 0 means y → 0 as y → 0

e^{sin2 y1 log}

y

sin y = e^{sin2 y1 (log}

y sin y−log 1)

by mean value Th , 1 < c < y

sin y → 1

= e^{sin2 y1 (1/c)(}

y sin y−1)

= e^(1/c)

y−sin y

sin3 y = e^(1/c)

y3 sin3 y

y−sin y y3

because 1/c→ 1 and y³

sin³y → 1 and y− sin y

y³ → 1/6 by taylor expansion

= e^1/6

(b)

x^xx = e^{ex log xlog x}− 1 x log x→ 0 and logx → −∞

e^{ex log xlog x}− 1 = e^−∞− 1 = −1

x^x → 1 x^xx = x¹ = 0 2. (12%, 6% each)

(a) Find d³ dx³

( x

√3

1 + x )

. (b) Find d

dx (

ln (cos⁻¹ 1

√x) )

.

Sol:

(a) by (f g)⁰⁰⁰ = f⁰⁰⁰g + 3f⁰⁰g⁰+ 3f⁰g⁰⁰+ f g⁰⁰⁰

3x⁰(1 + x)^−1/300+ x(1 + x)^−1/3000 = 3(−1/3)(−4/3)(1 + x)^−7/3+ x(−1/3)(−4/3)(−7/3)(1 + x)^10/3

= (4/3)(1 + x)^−7/3− (28/27)x(1 + x)^10/3

the original order is 1− 1/3 = 2/3, after three differentiation, it should be 2/3 − 3 = −7/3 (b) by chain rule

1 cos⁻¹1/√

x

√ −1

1− 1/x

−x^−3/2 2

3. (16%) Let H(x) =











 1

πtan⁻¹(ax + b

x), if x > 0,

c, if x = 0,

(1− ln 2^x)^1x, if x < 0.

(a) Find conditions of a, b, and c such that H is continuous. (6%) (b) Find conditions of a, b, and c such that H is differentiable. (10%) Sol:

(a)

lim

x→0⁻H(x) = lim

x→0⁻(1− ln 2^x)^1x

= lim

x→0⁻eln(1−x ln 2) x

= e^limx→0− ln(1−x ln 2) x

= e^{− ln 2} = 1 2

lim

x→0⁺H(x) = lim

x→0⁺

1

π tan⁻¹(ax + b x)

= 1

2 if b > 0

= 0 if b = 0

= −1

2 0 if b < 0 H(x) continuous at 0⇐⇒ , a ∈ R ,b > 0, c = 1

2

(b) lim

x→0− x− 0 = lim

x→0− x− 0 =−

4

xlim→0+

H(x)− H(0)

x− 0 = lim

x→0+

1

πtan⁻¹(ax +_x^b)− ¹₂

x = −1

πb

⇐⇒ b = 4 π(ln 2)²

H(x) differentiable at 0 when a∈ R, b = 4

π(ln 2)², c = 1 2

4. (14%) Find the line normal to the curve defined by tan⁻¹(xy) + ln (x + y) = x^y− 1 at (x, y) = (1, 0). Also find y⁰⁰ of the curve at (1, 0).

Sol:

tan⁻¹(xy) + ln(x + y) = x^y− 1 d tan⁻¹(xy)

dx + d ln(x + y)

dx = de^{y ln x} dx 1

(xy)²+ 1(y + xy⁰) + 1

x + y(1 + y⁰) = e^{y ln x}(y⁰ln x + y

x) (1)

Now we substitute x = 1, y = 0 into the equation:

1

0² + 1(0 + 1y⁰) + 1

1 + 0(1 + y⁰) = e^{0 ln 1}(y⁰ln 1 + 0 1) Hence we get: y⁰ =−1

2. And the line normal to that curve at (1, 0) is (y− 0) = −1

y⁰(x− 1), i.e., y = 2(x− 1).

Now taking implicit differentiation on (1) again, we get:

((xy)²+ 1)(y⁰+ y⁰+ xy⁰⁰)− (y + xy⁰)(2xy(y + xy⁰))

((xy)²+ 1)² + y⁰⁰(x + y)− (1 + y⁰)² (x + y)²

= e^{y ln x}(y⁰ln x + y

x)²+ e^{y ln x}(y⁰⁰ln x + y⁰

x + y⁰x− y x² ) Substitue x = 1, y = 0, y⁰ =−1

2 into it, we get:

f⁰⁰= 1 8

5. (14%) (a) Apply Generalized Mean Value Theorem to establish the inequalities

−1

3 < tan⁻¹x− x

x³ < −1

3(1 + x²), x > 0. (9%) (b) Use the result in (a) with x = 1

√3 to find an interval that contains π. Use the midpoint of this interval to estimate π. Also find the error of this approximation. (5%)

Sol:

(a) By Generalized MVT, x > 0,

tan⁻¹x− x

x³ =

1 1+c² − 1

3c² =− 1 3(1 + c²) for some c∈ (0, x).

Since 1 < 1 + c² < 1 + x²,

−1

3 <− 1

3(1 + c²) <− 1 3(1 + x²), we have the inequalities.

(b) tan⁻¹ 1

√3 = π 6, thus

−1 3 <

π 6 − ^√1₃

1 3√

3

<−1 4 16√

3

9 < π < 11√ 3 6 Midpoint approximation gives

π≈

√3 2

(16 9 +11

6 )

= 65 36

√3

Error is less than or equal to half length of interval

error≤

√3 2

(11 6 −16

9 )

=

√3 36

(Students can get 1% if error is estimated by total length of interval.)

6. (20%) Graph y = f (x) = (x + 2)e^x1. Be sure to write down the critical points, the intervals of monotonicity, the points of inflection, the intervals of concavity, and the asymptotes (if any).

Sol:

y = f (x) = (x + 2)e^1x, x6= 0

x→0lim⁺f (x) =∞ , lim

x→0⁻f (x) = 0 f⁰(x) = e^x1(x²− x − 2

x² ) = e^1x((x− 2)(x + 1)

x² )

critical points: x =−1, x = 2

intervals of increasing: (−∞, −1], [2, ∞)

f⁰⁰(x) = e^1x(5x + 2 x⁴ ) inflecion points: x = −2

5 intervals of concave up: (−2

5, 0), (0,∞) intervals of concave down: (−∞, −2

5) vertical asymptotes: x = 0,

xlim→∞(1 + 2

x)e^x1 = 1

xlim→∞(x + 2)e^1x − x = lim

x→∞x[(1 + 2

x)e^x1 − 1]

= lim

x→∞

(1 + ²_x)e^x1 − 1

1 x

= lim

x→∞

−_x²2e^x1 + (1 + ²_x)(−_x¹2)e^x1

−_x¹2

= lim

x→∞(2 + 1 + 2

x)e^1x = 3

x→−∞lim (1 + 2

x)e^x1 = 1

x→−∞lim (x + 2)e^1x − x = lim

x→−∞x[(1 + 2

x)e^x1 − 1]

= lim

x→−∞

(1 + ²_x)e^x1 − 1

1 x

= lim

x→−∞

−_x²²e^x1 + (1 + ²_x)(−_x¹²)e^1x

−_x¹²

= lim

x→−∞(2 + 1 + 2

x)e^1x = 3 asymptotes: y = x + 3

7. (10%) A man is in a boat 2 miles away from the nearest point on the coast. He is going to a point Q, 3 miles down the coast and 1 mile in land. If he can row 2 miles per hour, and walk 4 miles per hour, toward what point on the coast should he row in order to reach Q in the least time?

River 2 miles

1 mile 3 miles

Q

Sol:

River 2 miles

1 mile 3-x

Q x

f (x) = 2 + x

2 + 1 + (3− x) 4

= 1 4(2√

4 + x² +√

x²− 6x + 10)

f⁰(x) = 1

4( 2x

√4 + x² + 2x− 6 2√

x² − 6x + 10)

= 1

4( 2x

√4 + x² + √ x− 3 x²− 6x + 10) f⁰(x) = 0

⇒ 2x√

x²− 6x + 10 + (x − 3)√

4 + x² = 0 2x√

x²− 6x + 10 = −(x − 3)√ 4 + x² 4x²(x²− 6x + 10) = (x − 3)²(4 + x²)

4x⁴− 24x³+ 40x² = 4x² − 24x + 36 + x⁴− 6x³+ 9x² x⁴− 6x³+ 9x²+ 8x− 12 = 0

(x− 1)(x³− 5x²+ 4x− 12) = 0

The process below is to prove the equation is above zero.

x³− 4x²+ 4x− 12 − x² = x(x− 2)²+ 12− x² > 0 , for 0≤ x ≤ 3 and f⁰(1⁻) < 0 , f⁰(1) = 0 , f⁰(1⁺) > 0

x=1 is the only critical point in [0,3]

f (0) =

√4 2 +

√10 4 > 2 f (3) =

√13 2 + 1

4 > 2 f (1) is the least time:

f (1) =

√5 2 +

√5 4 = 3

4

√5 > 2 .