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98學年度第1學期 微積分甲二組期中考解答

1. (10%) Let f (x) = e^x· ln (2 + sin x).

(a) Find f⁰(x).

(b) Find f⁰(0).

Sol:

(a) f⁰(x) = e^xln (2 + sinx) + e^x cosx 2 + sinx (b) f⁰(0) = 1 · ln 2 + 1 ·1

2 = ln 2 +1 2 2. (10%) (a) Find lim

x→5

2^x− 32 x − 5 . (b) Find lim

x→−∞

√

x²+ 4x + 5 −√

x² + x + 1.

Sol:

(a)

x→5lim

2^x− 32 x − 5 = d

dx2^x|_x=5

or you can use the L’Hospital’s rule after identifying the limit is of the indeterminate form

0

0, Anyhow, the answer must have to do with the differentiation of 2^x at x = 5, which is d

dx2^x|_x=5 = 2^xln 2|_x=5 = 32 ln 2 (b) We first do some simplification:

√x²+ 4x + 5−√

x²+ x + 1 = (x²+ 4x + 5) − (x²+ x + 1)

√x²+ 4x + 5 +√

x²+ x + 1 = 3x + 4

√x²+ 4x + 5 +√

x²+ x + 1 Hence we have

x→−∞lim

√

x²+ 4x + 5 −√

x²+ x + 1 = lim

x→−∞

3 + _x⁴

√x²+4x+5

x +

√x²+x+1 x

= lim

x→−∞

3 + ⁴_x

−q

x²+4x+5 x² −q

x²+x+1 x²

= 3

−1 − 1 = 3

−2

3. (10%) Find lim

x→∞

sin(x⁻²+ e^−x) − sin(x⁻²)

e^−x .

Sol:

Method 1.

Let f (x) = sin x. By Mean Value Theorem,

f (sin(x⁻²+ e^−x) − f (x⁻²) = f⁰(c)e^−x, where c ∈ (x⁻², x⁻²+ e^−x).

So c → 0 as x → ∞. Therefore,

x→∞lim

sin(x⁻²+ e^−x) − sin(x⁻²)

e^−x = lim

c→0

f⁰(c)e^−x

e^−x = lim

c→0f⁰(c)

= lim

c→0cos(c⁻²) = cos 0 = 1.

Method 2.

x→∞lim

sin(x⁻²+ e^−x) − sin(x⁻²)

e^−x = lim

x→∞

sin x⁻²cos e^−x+ cos x⁻²sin e^−x− sin x⁻² e^−x

= lim

x→∞(sin x⁻²(cos e^−x− 1)

e^−x +sin e^−xcos x⁻² e^−x )

= ( lim

x→∞sin x⁻²)( lim

x→∞

cos e^−x− 1

e^−x ) + ( lim

x→∞cos x⁻²)( lim

x→∞

sin e^−x e^−x )

= 0 · 0 + 1 · 1 = 1.

4. (10%) For what values of a and b is the following equation true?

x→0lim

1 − cos x

x² + a + b x



= 0.

Sol:

By l’Hospital rule,we have

x→0lim

1 − cos x + ax²+ bx

x² = lim

x→0

sin x + 2ax + b 2x

if the RHS exists. Thus we may assume the RHS exits to try to find a and b Because the LHS exists,we have

x→0limsin x + 2ax + b = 0 ⇒ b = 0 Again by l’Hospital rule, we have

x→0lim

sin x + 2ax + b

2x = lim

x→0

cos x + 2a

2 = a +1 2

The above statement implies a = −1 2

5. (10%) A particle is moving along the curve y =√

x. As the particle passes through the point (4, 2), its x-coordinate increases at a rate 3 cm/s. How fast is the distance from the particle to the origin changing at this moment?

Sol:

Method 1.

(⁰= d dt) s =p

(x²+ y²), s⁰ = 2xx⁰+ 2yy⁰ 2px²+ y², y =√

x, y⁰ = x⁰ 2√

x, s⁰ = 2 · 4 + 2 · 2 · ¹

2√ 4

2√

16 + 4 · 3 = 27 4√

5cm/s Method 2.

s =p

(x²+ y²) = q

(x²+ (√

x)²) =p

(x²+ x) s⁰ = 2x + 1

2√ x²+ x

dx dt s⁰ = 2 · 4 + 1

2√

4²+ 4 · 3 = 27 4√

5cm/s

6. (15%) Find y⁰ and y⁰⁰ of the curve x²+ y² = (2x²+ 2y²− x)² at the point (0,1 2).

Sol:

implicit differentiation.

d

dx(x²+ y²) = d

dx(2x²+ 2y²− x)²

⇒ 2x + 2ydy

dx = (4x + 4ydy

dx − 1) · x · (2x²+ 2y²− x) (1) Plug in (x, y) = (0,1

2) dy dx   (0,¹₂)

= (2dy dx   (0,¹₂)

− 1) · 2 · (2 ·1

4) ⇒ dy dx   (0,¹₂)

= 1 Next, differentiate (1) w.r.t to x:

2+2(dy

dx)²+2yd²y

dx² = (4+4(dy

dx)²+4yd²y

dx²)·2·(2x²+2y²−x)+(4x+4ydy

dx−1)·2·(4x+4ydy dx−1)

Plug in







(x, y) = (0,¹₂)

dy dx

(0,¹₂) = 1

⇒ 2 + 2 + 1 · d²y dx²    (0,¹₂)

= (4 + 4 + 2 d²y dx²    (0,¹₂)

) · 2 · (2 · 1

4) + (2 · 1 − 1) · 2 · (2 · 1 − 1) 4 + d²y

dx²    (0,¹₂)

= (8 + 2 d²y dx²    (0,¹₂)

) + 2 ⇒ d²y dx²    (0,¹₂)

= −6

7. (15%) Consider the function f (x) = x²+ 1

√x²− 4, for x < −2 and x > 2.

(a) Find the intervals of increase and decrease. (b) Find the maximum and the minimum.

(c) Find the asymptotes.

Sol:

(a), (b) f⁰(x) =

√x²− 4 · (2x) − (x²+ 1) · ^2x

2√ x²−4

x²− 4 = x(x − 3)(x + 3) (x²− 4)³² Increasing interval: [−3, −2) and [3, ∞).

Descending interval: (−∞, −3] and (−2, 3].

So f (x) only has minima at x = −3, 3. =⇒ Minimum: f (−3) = f (3) = 2√ 5.

(c) There are four asymptotes to the function. As x approach ±2, f(x) go to infinite. Hence x = ±2 are asymptotes. Moreover, to see if there exists slant asymptotes,i.e mx + b, we need to observe

m = lim

x→∞

x²+ 1 x√

x²− 4 = lim

x→∞

1 + _x¹2

q 1 − _x⁴2

= 1

b = lim

x→∞f (x) − x

= lim

x→∞

x²+ 1 − x√ x²− 4

√x²− 4

= lim

x→∞

(x² + 1) − (x√

x² − 4)

√x²− 4

(x² + 1) + (x√

x²− 4) (x² + 1) + (x√

x²− 4)

= lim

x→∞

6x²+ 1

√x²− 4(x²+ 1 + x√

x²− 4) = 0

By definition of asymptote, y = 1 · x + 0 is an asymptote. Since f(x)=f(-x), y=-x is also an asymptote too.

8. (10%) Evaluate lim

n→∞

1 n

"

ln ⁿ⁺¹_n

n+1 n

+ln ⁿ⁺²_n

n+2 n

+ · · · + ln ⁿ⁺ⁿ_n

n+n n

# . Sol:

n→∞lim 1 n

n

X

k=1

ln(1 +_n^k) 1 + ^k_n =

Z 2 x=1

ln x x dx

= Z ln 2

u=0

udu

where u = ln x, = 1 2(ln 2)²

9. (10%) Find the derivative of h(x) = Z sin x

0

√

1 + r³dr.

Sol:

h(x) = Z sin x

0

√

1 + r³dr Let G(x) =

Z x 0

√

1 + r³dr d

dxG(x) =√ 1 + x³ h(x) = G(sin x) d

dxh(x) = d

d sin xG(sin x)d sin x dx =p

1 + sin³x cos x