(1) (28 Points) Evaluate (a) lim
x→π6
2 sin2x + sin x − 1 2 sin2x − 3 sin x + 1. Solution:
For x 6= π
6, we have
2 sin2x + sin x − 1
2 sin2x − 3 sin x + 1 =(2 sin x − 1)(sin x + 1) (2 sin x − 1)(sin x − 1)
=sin x + 1 sin x − 1. Hence
lim
x→π6
2 sin2x + sin x − 1
2 sin2x − 3 sin x + 1 = lim
x→π6
sin x + 1
sin x − 1 = sinπ6 + 1 sinπ6 − 1 =
1 2+ 1
1
2− 1 = −3.
(b) lim
x→0
1 − cos x x2 . Solution:
For x 6= 0,
1 − cos x
x2 = 2 sin2 x2 x
= 1 2
sinx2
x 2
2 .
Here we use 1 − cos x = 2 sin2x
2. Since lim
θ→0
sin θ
θ , we find
x→0lim
1 − cos x x2 = lim
x→0
1 2
sinx2
x 2
2
=1
2 · 12=1 2. (c) lim
x→0
tan 3x sin 8x. Solution:
For x 6= 0, we write tan 3x
sin 8x = 8x
sin 8x·sin 3x 3x · 1
cos 3x·3 8. Since lim
x→0
sin θ
θ = cos θ = 1, we find
x→0lim tan 3x sin 8x = lim
x→0
8x
sin 8x· sin 3x 3x · 1
cos 3x·3 8
= 3
8· 1 · 1 · 1 = 3 8.
1
(d) lim
x→1
x + x2+ · · · + xn− n
x − 1 .
Solution:
Write xk− 1 = (x − 1)(xk−1+ · · · + x + 1). Hence
x + x2+ · · · + xn− n =
n
X
k=1
xk
!
− n
=
n
X
k=1
(xk− 1)
=
n
X
k=1
(x − 1)(xk−1+ · · · + x + 1)
= (x − 1)
n
X
k=1
(xk−1+ · · · + x + 1).
For x 6= 1,
x + x2+ · · · + xn− n
x − 1 =
n
X
k=1
(xk−1+ · · · + x + 1).
By properties of limits and the continuity of polynomials, we have
x→1lim
x + x2+ · · · + xn− n
x − 1 =
n
X
k=1
k = n(n + 1)
2 .
(e) lim
x→0
√1 + tan x −√
1 + sin x
x3 .
Solution:
Using the formula A2− B2= (A + B)(A − B), we write
√1 + tan x −√
1 + sin x = tan x − sin x
√1 + tan x +√
1 + sin x
= 1
√1 + tan x +√
1 + sin x
sin x cos x− sin x
= 1
√1 + tan x +√
1 + sin x· sin x
cos x(1 − cos x).
Hence we know
√1 + tan x −√
1 + sin x
x3 = 1
√1 + tan x +√
1 + sin x·sin x x · 1
cos x·1 − cos x x2 .
Note that lim
x→0
√ 1
1 + tan x +√
1 + sin x = 1
2, and lim
x→0
sin x
x = 1 and lim
x→0
1
cos x = 1 and
x→0lim
1 − cos x x2 = 1
2, we obtain
x→0lim
√1 + tan x −√
1 + sin x
x3 = 1
2· 1 · 1 ·1 2 = 1
4. (f) lim
x→∞(p3
x3+ 3x2−p
x2− 2x).
Solution:
x We use the formula A6− B6= (A − B)(A5+ A4B + A3B2+ A2B3+ AB4+ B5).
Write p3
x3+ 3x2−p
x2− 2x = ((x3+ 3x2)2− (x2− 2x)3)
· {(x3+ 3x2)53 + (x3+ 3x2)43(x2− 2x)12
+ (x3+ 3x2)43(x2− 2x)22 + (x3+ 3x2)23(x2− 2x)32 + (x3+ 3x2)13(x2− 2x)42 + (x2− 2x)52}−1
= (12 − 3 x+ 8
x2)
· {(1 + 3
x)53 + (1 + 3
x)43(1 − 2 x)12 + (1 + 3
x)33(1 − 2
x)22 + (1 + 3
x)23(1 − 2 x)32 + (1 + 3
x)13(1 − 2
x)42 + (1 − 2 x)52}−1. Hence
x→∞lim(p3
x3+ 3x2−p
x2− 2x) = 12 6 = 2.
(g) lim
x→π2cos
2x + sin(3π 2 + x)
. Solution:
lim
x→π2cos
2x + sin(3π 2 + x)
= cos
lim
x→π2 2x + lim
x→π2sin(3π 2 + x)
= cos(2 ·π
2 + sin 2π)
= cos π = −1.
(2) (12 Points) Evaluate (a)
Z
√ 3 2
√1 2
p1 − x2dx.
Solution: The integral is the area of the region DEIH. We write
Z
√3 2
√1 2
p1 − x2dx = Z
√3 2
0
p1 − x2dx − Z √1
2
0
p1 − x2dx.
We know Z
√ 3 2
0
p1 − x2dx equals the area of the region AEIC and Z √12
0
p1 − x2dx equals the area of the region ADHC.
Since the area of the region EBI is given by π
12 −1 2 ·
√3 2 ·1
2 (area of ∆AEI) = π 12−
√3 8 , the area of the region AEIH is given by
π 4 − π
12−
√3 8
!
=π 6 +
√3 8 . In other words,
Z
√ 3 2
0
p1 − x2dx = π 6 +
√3 8 .
Similarly, we know the area of the region DBH is given by π
8 −1 2 · 1
√2· 1
√2(area of ∆ADH) = π 8 −1
4. Hence the area of the region ADHC is given by
π 4 − π
8 −1 4
= π 8 +1
4.
In other words,
Z √1
2
0
p1 − x2dx = π 8 +1
4. We conclude that
Z
√ 3 2
√1 2
p1 − x2dx = π 6 +
√3 8
!
− π 8 +1
4
= π 24+
√3 − 2 8 .
(b) Z 6
3
[x]dx. Here [x] is the Gauss-symbol of x.
Solution:
The integral of f (x) over [3, 6] is the sum of all areas of the region I, II, II. Hence Z 6
3
[x]dx = 1 · 3 + 1 · 4 + 1 · 5 = 12.
(c) Z 2
−1
f (x)dx. Here
f (x) =
2x + 2 if −1 ≤ x ≤ −1/2,
−4x − 1 if −1/2 < x ≤ 0, x − 1 if 0 < x ≤ 1, 1 if 1 < x ≤ 2.
Solution:
The integral Z 2
−1
f (x)dx equals the sum of areas of I and III minus the area of II, i.e.
Z 2
−1
f (x)dx = A(I) − A(II) + A(III).
The area of I, II, II are given by A(I) = 1
2 ·3 4 · 1 = 3
8(area of ∆ABH) A(II) = 1
2 ·5 4 · 1 = 5
8(area of ∆CDH) A(III) = 1 · 1(area of the square DGF E).
Hence
Z 2
−1
f (x)dx = 3 8−5
8 + 1 = 3 4.
