(a) The degree of the numerator is greater than that of the denominator, so the limit does not exist

Solutions for Calculus Midterm

1. (a) The degree of the numerator is greater than that of the denominator, so the limit does not exist. And 2 + x²

1 − x behaves as x²

−x = −x when x → −∞.

Thus lim

x→−∞

2 + x² 1 − x = ∞.

(b) We know that −1 ≤ sin x ≤ 1.

Then

− 1

|x| ≤ sin x x ≤ 1

|x|. Also

x→∞lim − 1

|x| = lim

x→∞

1

|x| = 0.

By Sandwich Theorem, we obtain that

x→∞lim sin x

x = 0.

(c) lim

h→0

a^h− 1

h = da^x

dx |^x=0 = a^xln a|^x=0 = a⁰ln a = ln a.

2. (a) Suppose a is a fixed point of {aⁿ}. Then a = 1

2(a +4 a).

Hence a = ±2.

That is the fixed points of {aⁿ} are 2, −2.

(b) Since a0 = 1 > 0, then

a_n+1 = 1

2(an+ 4

a_n) > 0 ∀n ∈ N.

Thus limn→∞a_n = 2.

3. (a) f^′(x) = cos√

3x² + 3x · ¹₂(3x² + 3x)⁻¹² · (6x + 3) (b) f (x) = ln(_x¹) = − ln x

df

dx = −¹x = −x⁻¹

d²

dx²f = _dx^d(−x⁻¹) = x⁻²

d³

dx³f = _dx^d(x⁻²) = −2x⁻³ .

. .

Then we find that _dx^dnnf(x) = (−1)ⁿ(n − 1)!x⁻ⁿ. (c)

f(x) = x^aa^−x

f^′(x) = ax^a−1a^−x+ x^aa^−x(−1) ln a

= x^a−1a^−x(a − x ln a) f^′( a

ln a) = ( a

ln a)^a−1a^{−ln aa} (a − a

ln aln a) = 0

(d) f (x) = e^−x

2 2 + 2x And f (0) = 1.

f^′(x) = −xe^−x

2 2 + 2.

Then

d

dxf⁻¹(1) = (f⁻¹)^′(1)

= (f⁻¹)^′(f (0))

= 1

f^′(f⁻¹(f (0)))

= 1

f^′(0)

= 1 2 4. (a) _dx^d(^x₂₅²) − dx^d(^y₉²) = _dx^d(1) = 0

⇒ 25²x− ²9y_dx^dy = 0 Thus _dx^dy = _25y^9x

dy

dx|(²⁵₃,4) = 3 4.

Thus the slope of the normal line is −⁴3. The normal line is y = 4 − ⁴₃(x − ²⁵₃).

(b)

d²y

dx² = d dx( 9x

25y)

= 9

25(y−dx^dyx y² )

= 9

25(y−_25y^9x x y² )

= 225y²− 81x² 625y³ d²y

dx²|(²⁵₃,4) = 225 × 4²− 81 × (²⁵3)² 625 × 4³

= −81 1600 5. (a)

f^′(0) = lim

h→0

f(0 + h) − f(0) h

= lim

h→0

f(h) h

=







h→0lim h

h = 1 if h = ¹_n

h→0lim

−h

h = −1 if h 6= ¹n

Thus f^′(0) does not exist.

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(b)

f^′(0) = lim

h→0

f(0 + h) − f(0) h

= lim

h→0

f(h) h

=







h→0lim h²

h = lim

h→0h= 0 if h = _n¹

h→0lim

−h²

h = lim

h→0−h = 0 if h 6= n¹

Thus f^′(0) = 0.

(c) f^′(0) = lim

h→0

f(0 + h) − f(0)

h = lim

h→0

f(h) Consider h

h→0lim⁺ f(h)

h = lim

h→0⁺

e^h− 1

h = (e^h)^′|^h=0 = e^h|^h=0 = 1

h→0lim⁻ f(h)

h = lim

h→0⁻

h²− 1

h = (h²)^′|^h=0 = (2h)|^h=0 = 0 Then f^′(0) = lim

h→0

f(h)

h does not exist.

6. (a) (b) (c)

dS

dt = 1.162 × 0.933B^0.933−1dB dt 1

S dS

dt = 1

1.162B^0.9331.162 × 0.933B^−0.067dB dt

= 0.9331 B

dB dt Thus 1

S dS

dt ≤ 1 B

dB dt. (d) From (c), dS

dt = 1.084 1 B^0.067

dB dt. When 1.084

B^0.067 ≤ 1, dS

dt ≤ dB dt. Solve 1.084

B^0.067 = 1, we obtain B = 3.34(cm).

(e) S = 1.162B^0.933 and B(t) =√ 300t.

We solve√

300t = 1.162(√

300t)^0.933, then we get t = 1

300(1.162)^0.0672 ≈ 0.2947(month).

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