Math 110,
29/5/1431H
Test 2–Version B
1
Instructions. (30 points) Solve each of the following problems.
(1pts )
𝑥4 − 1
=
𝑥→−1 𝑥 − 1
(a) −4
1. lim
(b) Does Not Exist
(c) 4
Solution:
(d)
 0
(−1)4 − 1
𝑥4 − 1
=
𝑥→−1 𝑥 − 1
−1 − 1
1−1
=
−2
0
=0
=
−2
lim
(1pts )
2. If cos 𝑥 =
−3
4
(a)
3
,
5
3𝜋
< 𝑥 < 2𝜋 then sin 𝑥 =
2
−5
4
Solution:
(c)
(b)
−4
3
(d)

−4
5
adjacent
3
=
we draw a triangle is shown below:
5
hypotenuse
5
3𝜋
< 𝑥 < 2𝜋,
Now, 𝐻 2 = 𝐴2 + 𝑂2 ⇒ 25 = 9 + 𝑂2 ⇒ 𝑂 = 4. Since
2
then 𝑥 lies in the forth quadrant. Hence since the 𝑦− axis is 𝑥
3
negative, then sin 𝑥 < 0 and since the 𝑥−axis is positive then
−4
.
cos 𝑥 > 0. Hence sin 𝑥 =
5
Since cos 𝑥 =
(1pts )
𝑥 − 4𝑥3
=
𝑥→∞ 2𝑥3 − 1
−1
(a)
2
4
3. lim
(b) ∞
(c) 0
(d)
 −2
Solution:
Since lim (𝑥 − 4𝑥3 ) = −∞, lim (2𝑥3 − 1) = ∞ we have I.F. type −∞/∞. Divide
𝑥→∞
𝑥→∞
each term in the numerator and each term in the denominator by the highest power in the
Math 110,
29/5/1431H
Test 2–Version B
2
denominator.
𝑥 − 4𝑥3
𝑥−
𝑥3
= lim
lim
3
3
𝑥→∞ 2𝑥 − 1
𝑥→∞ 2𝑥 + 1
𝑥3
𝑥
4𝑥3
−
3
3
= lim 𝑥 3 𝑥
𝑥→∞ 2𝑥
1
− 3
3
𝑥
𝑥
1
−4
2
= lim 𝑥
1
𝑥→∞
2− 3
𝑥
0−4
=
= −2
2−0+0
4𝑥3
)
( )
(𝜋)
(𝜋)
7𝜋
7𝜋
cos
− sin
cos
=
4. sin
18
18
18
18
1
(b) 2
(a)
2
√
3
(d) 0
(c)

2
Solution:
(
(1
pts
)
(
sin
(1pts )
(c) 𝑥 = 0,
Solution:
)
( )
(
)
(𝜋)
(𝜋)
7𝜋
7𝜋
𝜋
cos
− sin
cos
= sin
−
18
18
18
18
18
(𝜋 )
== sin
3
√
3
.
=
2
𝑥2 − 2𝑥 − 3
𝑥3 − 9𝑥
𝑥 = ±3
5. The curve 𝑓 (𝑥) =
(a) 𝑥 = 0,
7𝜋
18
has a vertical asymptote at
𝑥=3
(b) 𝑦 = 0,
𝑦 = −3
(d)
 𝑥 = 0,
𝑥 = −3
(𝑥 − 3)(𝑥 + 1)
. The zeroes of the denominator are −3, 0, and 3. To check
𝑥(𝑥 − 3)(𝑥 + 3)
that 𝑥 = 3 is a vertical asymptote or not we take the limit at 3 from both sides. lim 𝑓 (𝑥) =
Write 𝑓 (𝑥) =
𝑥→3
+ 1)
4
2
(𝑥
−3)(𝑥
𝑥+1
= lim
=
= . Hence 𝑥 = 3 is not a vertical asymptote.
lim
𝑥→3 𝑥
𝑥→3 𝑥(𝑥 + 3)
(𝑥
− 3)(𝑥 + 3)
18
9
+
(𝑥 − 3)(𝑥 + 1)
= ∞. Hence
+
𝑥(𝑥 − 3)(𝑥 + 3)
𝑥 = −3 is a vertical asymptote. To check that 𝑥 = 0 we take the limit lim 𝑓 (𝑥) =
To check that 𝑥 = −3 we take the limit
lim 𝑓 (𝑥) = lim
𝑥→−3+
𝑥→−3+
𝑥→0+
Math 110,
29/5/1431H
Test 2–Version B
3
−
(𝑥 − 3)(𝑥 + 1)
lim
−
𝑥→0+
= ∞. Hence 𝑥 = 0 is a vertical asymptote. Thus the function has
𝑥(𝑥 − 3)(𝑥 + 3)
vertical asymptote at 𝑥 = 0, and 𝑥 = −3.
(1pts )
6. If
⎧
𝑥 + 2,
if 𝑥 < −3;


⎨
2,
if 𝑥 = −3;
Then
𝑓 (𝑥) =

𝑥3 + 27

⎩
,
if
𝑥
>
−3.
𝑥2 − 9
9
(a)
2
(c) 0
lim 𝑓 (𝑥) =
𝑥→−3+
(b) −
3
2
(d)
 −
9
2
Solution:
Note that when computing limit as 𝑥 → 𝑎+ means you approaches 𝑎 from the right side
that is 𝑥 > 𝑎.
𝑥3 + 27
𝑥→−3+ 𝑥2 − 9
𝑥3 + 27
A direct substation will give us I.F. 0/0
= lim
𝑥→−3+ 𝑥2 − 9
(𝑥 + 3)(𝑥2 − 3𝑥 + 9)
= lim
Factoring 𝑥 + 3 from denominator and numerator
(𝑥 + 3)(𝑥 − 3)
𝑥→−3+
2 − 3𝑥 + 9)
(𝑥
+3)(𝑥
= lim
− 3)
(𝑥
+3)(𝑥
𝑥→−3+
𝑥2 − 3𝑥 + 9
= lim
𝑥−3
𝑥→−3+
2
(−3) − 3(−3) + 9
=
−3 − 3
9
27
=− .
=
−6
2
lim 𝑓 (𝑥) = lim
𝑥→−3+
(1pts )
7.
𝑑42
(cos 𝑥) =
𝑑𝑥42
(a) − sin 𝑥
(b) sin 𝑥
(c) cos 𝑥
Solution:
(d)
 − cos 𝑥
Since 42 = 4(10) + 2 then
(1pts )
𝑥2 + 2𝑥
≤ 𝑓 (𝑥) ≤ 3𝑥 + 2,
𝑥
(a) 0
8. If
𝑑2
𝑑
𝑑42
(− sin 𝑥) = − cos 𝑥.
(cos
𝑥)
=
(cos 𝑥) =
42
2
𝑑𝑥
𝑑𝑥
𝑑𝑥
𝑥 ∈ [−1, 1], 𝑥 ∕= 0,
(c) −2
Solution:
then lim 𝑓 (𝑥) =
𝑥→0
(b)
 2
(d) 1
𝑥2 + 2𝑥
𝑥(𝑥 + 2)
= lim
= lim (𝑥 + 2) = 2, and lim (3𝑥 + 2) = 2, then by The
𝑥→0
𝑥→0
𝑥→0
𝑥→0
𝑥
2𝑥
Sandwich Theorem we have lim 𝑓 (𝑥) = 2.
Since lim
𝑥→0
Math 110,
(1pts )
29/5/1431H
Test 2–Version B
tan3 (2𝑥)
=
𝑥→0
𝑥3
1
(a)
8
9. lim
(b) 2
(c)
 8
(d)
Solution:
tan3 (2𝑥)
= lim
lim
𝑥→0
𝑥→0
𝑥3
= lim
𝑥→0
(
=
(
(
tan (2𝑥)
𝑥
)3
2 tan (2𝑥)
2𝑥
2𝑥
2 lim
𝑥→0 tan (2𝑥)
)3
1
2
( 𝑎 )𝑛
𝑎𝑛
=
.
𝑏𝑛
𝑏
make the top similar to the angle
)3
Use that lim
𝑥→0
tan 𝑥
𝑥
= 1 = lim
𝑥→0 tan 𝑥
𝑥
= (2.1)3 = 8.
(1pts )
tan (2𝑡) sin 𝑡
=
𝑡→0
𝑡2
1
(a)
2
10. lim
(b)
 2
(c) 1
(d) −2
Solution:
Direct substation will give us I.F. type 0/0.
lim
𝑡→0
(1pts )
tan (2𝑡) sin 𝑡
tan (2𝑡) sin 𝑡
= lim
𝑡→0
𝑡2
𝑡
𝑡
tan (2𝑡) sin 𝑡
= 2 lim
𝑡→0
2𝑡
𝑡
= 2(1)(1) = 2
cos 𝑥
, then 𝑦 ′ =
1 − sin 𝑥
sin 𝑥
(a)
1 − sin 𝑥
tan 𝜃
sin 𝜃
= 1 = lim
𝜃→0 𝜃
𝜃→0 𝜃
use lim
11. If 𝑦 =
− cos 𝑥
(1 − cos 𝑥)2
Solution:
(c)
(b)

1
1 − sin 𝑥
(d)
cos 𝑥
1 − sin 𝑥
4
Math 110,
29/5/1431H
𝑦′ =
=
=
=
=
(1pts )
Test 2–Version B
(1 − sin 𝑥)(− sin 𝑥) − (cos 𝑥)(− cos 𝑥)
(1 − sin 𝑥)2
− sin 𝑥 + sin 𝑥 sin 𝑥 + cos 𝑥 cos 𝑥
(1 − sin 𝑥)2
− sin 𝑥 + sin2 𝑥 + cos2 𝑥
(1 − sin 𝑥)2
1 − sin 𝑥
(1 − sin 𝑥)2
1
.
1 − sin 𝑥
5
Use the fact cos2 𝑥 + sin2 𝑥 = 1
simplify
12. The graph of the function 𝐹 (𝑥) = 𝑥3 − 27𝑥, has a horizontal tangent line at
(a)
 𝑥 = ±3
(b) 𝑥 = 3
(c) 𝑥 = −3
Solution:
(d) 𝑥 = 0
𝐹 (𝑥) = 𝑥3 − 27𝑥
𝐹 ′ (𝑥) = 3𝑥2 − 27 = 3(𝑥2 − 9)
𝐹 ′ (𝑥) = 0
3(𝑥2 − 9) = 0
𝑥 = ±3.
(1pts )
13. An equation for the tangent line to the curve 𝑓 (𝑥) = 𝑥3 + 𝑥
(a) 𝑦 = −4𝑥 − 2
(b)
 𝑦 = 4𝑥 − 2
at 𝑥 = 1 is
(c) 𝑦 = 4𝑥 + 2
(d) 𝑦 = −4𝑥 + 6
Solution:
The slope of the tangent line to 𝑓 (𝑥) = 𝑥3 + 𝑥 at 𝑥 = 1 is 𝑓 ′ (1).
𝑓 (𝑥) = 𝑥3 + 𝑥 ⇒ 𝑓 ′ (𝑥) = 3𝑥2 + 1.
Hence
the slope of the tangent = 𝑓 ′ (1) = 4.
Also 𝑓 (1) = (1)3 + (1) = 2. Now, we have 𝑚 = 4 and (1, 2), hence
𝑦 − 𝑦1 = 𝑚(𝑥 − 𝑥1 ) ⇒ 𝑦 − 2 = 4(𝑥 − 1) ⇒ 𝑦 − 2 = 4𝑥 − 4 ⇒ 𝑦 = 4𝑥 − 2.
Math 110,
(1pts )
29/5/1431H
Test 2–Version B
6
14. The accompanying figure shows the
graph of 𝑦 = 𝑓 (𝑥). Then the period of
𝑦 = 𝑓 (𝑥) is
𝑦
(a) 1
𝑦 = 𝑓 (𝑥)
1
(b) 4
-4
-3
-2
-1
1
2
𝑥
3
-1
(c) 2𝜋
-2
(d)
 2
Solution:
From the graph we can see that the function repeat itself every 2 units. Hence the
period is 2
(1pts )
15. If 𝑓 (1) = 3, then lim 𝑓 (𝑥) must exist.
𝑥→1
(a) True
(b)
 False
Solution:
⎧
⎨ 1,
False. Let 𝑓 (𝑥) =
3,
⎩
−1,
lim 𝑓 (𝑥), Hence lim 𝑓 (𝑥) does
𝑥→1
𝑥→1−
(1pts )
if𝑥 > 1;
if𝑥 = 1; Then 𝑓 (1) = 3, but lim 𝑓 (𝑥) = 1 ∕= −1 =
𝑥→1+
if𝑥 < 1.
not exist.
16. The accompanying figure shows the graph of 𝑦 =
𝑓 (𝑥). Then 𝑓 ′ (−2) < 𝑓 ′ (0).
𝑦
(a)
 True
𝑦 = 𝑓 (𝑥)
1
(b) False
-4
-3
-2
-1
1
2
3
𝑥
-1
-2
Solution:
From the graph we can see that the tangent line to the graph of 𝑦 = 𝑓 (𝑥) at −2 is
falling down (left to right) and hence its slope is negative. Thus, since the first derivative
is the slope of the tangent line to the graph at −2 then 𝑓 ′ (−2) < 0.
From the graph we can see that the tangent line to the graph of 𝑦 = 𝑓 (𝑥) at 0 is rising
up (left to right) and hence its slope is positive. Thus, since the first derivative is the slope
of the tangent line to the graph at 0 then 𝑓 ′ (0) > 0. Now, 𝑓 ′ (−2) < 0 < 𝑓 ′ (0).
(1pts )
17. The accompanying figure shows the graph of 𝑦 =
𝑓 (𝑥). Then lim 𝑓 (𝑥) =
𝑥→0
(a) 0
(c) Does Not Exist
𝑦
(b)
 1
1
(d) −1
-4
-3
-2
-1
𝑦 = 𝑓 (𝑥)
1
-1
-2
2
3
𝑥
Math 110,
29/5/1431H
Test 2–Version B
7
Solution:
since lim 𝑓 (𝑥) = 1 = lim 𝑓 (𝑥), then lim 𝑓 (𝑥) = 1
𝑥→0−
(1pts )
𝑥→0+
𝑥→0
𝑥−3
√
=
𝑥→3 2 − 𝑥 + 1
18. lim
(a) 4
(b)
−1
4
(c) 0
(d)
 −4
Solution:
A direct substation will give us I.F. 0/0. Now,
√
2+ 𝑥+1
𝑥−3
𝑥−3
√
√
√
= lim
⋅
lim
𝑥→3 2 − 𝑥 + 1
𝑥→3 2 − 𝑥 + 1 2 + 𝑥 + 1
√
(𝑥 − 3)(2 + 𝑥 + 1)
√
= lim
𝑥→3
22 − ( 𝑥 + 1)2
√
(𝑥 − 3)(2 + 𝑥 + 1)
= lim
𝑥→3
4 − (𝑥 + 1)
√𝑥 + 6 + 3)
(𝑥
− 3)(
= lim
𝑥→3
−
(𝑥
−3)
√
= lim [−(2 + 𝑥 + 1)]
𝑥→3
√
= −[2 + 3 + 1]
Multiply by 1 =
√
2+ 𝑥+1
√
2+ 𝑥+1
Simplify
= −4.
(1pts )
𝑦
19. The accompanying figure shows the graph of 𝑦 =
𝑓 (𝑥). Then 𝑓 is differentiable at 𝑥 = −1.
4
(a) True
3
(b)
 False
𝑦 = 𝑓 (𝑥)
2
1
-4
-3
-2
-1
1
-1
Solution:
𝑓 is not differentiable at 𝑥 = −1 since the the function is discontinuous at 𝑥 = −1.
(1pts )
20. If 𝑦 = 𝑥 sin 𝑥 csc 𝑥,
(a)
 1
then 𝑦 ′ =
(b) −1
(c) 0
Solution:
(d) − cos 𝑥 csc 𝑥 cot 𝑥
𝑦 = 𝑥 sin 𝑥 csc 𝑥
1
𝑦 = 𝑥 sin 𝑥
sin 𝑥
𝑦=𝑥
𝑦 ′ = 1.
2
𝑥
Math 110,
(1pts )
29/5/1431H
21. The function 𝑓 (𝑥) =
(a) (−∞, −2] ∪ [2, ∞)
(c) (−∞, 2) ∪ (2, ∞)
√
Test 2–Version B
2 − ∣𝑥∣
8
is continuous on
(b)
 [−2, 2]
(d) [2, ∞)
√
Solution: Notice that 𝑓 (𝑥) = 2 − ∣𝑥∣ is an even root function, then it is continuous on its
domain which is the set of real number such that 2 − ∣𝑥∣ ≥ 0. Now
2 − ∣𝑥∣ ≥ 0 ⇔ −∣𝑥∣ ≥ −2
⇔ ∣𝑥∣ ≤ 2
⇔ −2 ≤ 𝑥 ≤ 2.
Hence 𝑓 is continuous on [−2, 2].
(1pts )
22. The accompanying figure shows the graph of 𝑦 =
𝑓 (𝑥). Then 𝑓 ′ (1) =
𝑦
(a)
 0
𝑦 = 𝑓 (𝑥)
1
(b) −1
-4
(c) Undefined
-3
-2
-1
1
2
-1
(d) 1
-2
Solution:
From the graph we can see that the tangent line to the graph of 𝑦 = 𝑓 (𝑥) at 1 is
horizontal (𝑦 = 1) and hence its slope is zero. Now, since the first derivative is the slope of
the tangent line to the graph at 2 then 𝑓 ′ (1) = 0.
(1pts )
23. If 𝑦 = 𝑥2 +
(a)
√
1
−
5, then 𝑦 ′′′ =
𝑥3
1
60
+ √
6
𝑥
2 5
60
𝑥6
Solution:
(c)
(b)

−60
𝑥6
(d)
−60
1
− √
6
𝑥
2 5
√
1
− 5
3
𝑥
√
2
𝑦 = 𝑥 + 𝑥−3 − 5
𝑦 = 𝑥2 +
𝑦 ′ = 2𝑥 − 3𝑥−4
𝑦 ′′ = 2 + 12𝑥−5
𝑦 ′′′ = −60𝑥−6 =
−60
.
𝑥6
3
𝑥
Math 110,
(1pts )
29/5/1431H
Test 2–Version B
3𝑥 + 1
, then 𝑦 ′ =
𝑥−1
2
(a)
(𝑥 − 1)2
24. If 𝑦 =
6𝑥 − 4
(𝑥 − 1)2
Solution:
(c)
(b)
−2
(𝑥 − 1)2
(d)

−4
(𝑥 − 1)2
Bottom Derivative of Top
′
Top
(𝑥 − 1)
(3𝑥 + 1)′
− (3𝑥 + 1)
𝑦 =
Derivative of Bottom
(𝑥 − 1)′
Bottom2
(𝑥 − 1)2
(𝑥 − 1)(3) − (3𝑥 + 1)(1)
(𝑥 − 1)2
3𝑥 − 3 − (3𝑥 + 1)
=
(𝑥 − 1)2
3𝑥 − 3 − 3𝑥 − 1)
=
(𝑥 − 1)2
−4
=
.
(𝑥 − 1)2
=
(
(1
pts
)
25. lim cos
𝜃→0
𝜋𝜃
sin (𝜃)
)
=
(a)
 −1
(b) 0
(c) 1
Solution:
(d) ∞
(
lim cos
𝜃→0
𝜋𝜃
sin (𝜃)
)
(
)
𝜋𝜃
= cos lim
𝜃→0 sin (𝜃)
)
(
𝜋𝜃
= cos
sin (𝜃)
= cos (𝜋)
= −1.
(1pts )
26. If 𝑦 = 𝑥 sin 𝑥, then 𝑦 ′ =
(a) cos 𝑥
(c) sin 𝑥 − 𝑥 cos 𝑥
Solution:
(b) sin 𝑥 + 1
(d)
 sin 𝑥 + 𝑥 cos 𝑥
𝑦 ′ = (𝑥)′ sin 𝑥 + 𝑥(sin 𝑥)′
= sin 𝑥 + 𝑥(cos 𝑥)
= sin 𝑥 + 𝑥 cos 𝑥.
9
Math 110,
(1pts )
29/5/1431H
Test 2–Version B
𝑥2 − 9
=
𝑥→3 𝑥2 − 𝑥 − 6
5
(a)
6
27. lim
(b)

−6
5
Solution:
(c)
𝑥2 − 9
𝑥2 − 9
=
lim
𝑥→3 𝑥2 − 𝑥 − 6
𝑥→3 𝑥2 − 𝑥 − 6
(𝑥 + 3)(𝑥 − 3)
= lim
𝑥→3 (𝑥 − 3)(𝑥 + 2)
(𝑥 + 3)
(𝑥
−3)
= lim
+ 2)
𝑥→3 (𝑥
−3)(𝑥
𝑥+3
= lim
𝑥→3 𝑥 + 2
6
3+3
= .
=
3+2
5
28.
lim
𝑥→−2−
6
5
(d) 0
lim
(1pts )
10
A direct substation will give us I.F. 0/0
Factoring 𝑥 − 3 from denominator and numerator
𝑥2 − 1
=
4𝑥 + 8
(a) ∞
(b) 0
(c) Does Not Exist
Solution:
(d)
 −∞
lim
𝑥→−2−
𝑥2 − 1
𝑥2 − 1
= lim
4𝑥 + 8 𝑥→−2− 2𝑥 + 4
−𝑥 − 1
= lim
−
𝑥+2
𝑥→−2
A direct substation gives I.F. 3/0.
If 𝑥 < −2 and near −2,then 𝑥2 − 1 > 0, 4𝑥 + 8 < 0.
+
= lim
𝑥2 − 1
𝑥→−2−
−
4𝑥 + 8
= −∞
𝑦
𝑦=
3
𝑥2 − 1
2𝑥 + 4
2
1
-8
-7
-6
-5
-4
-3
-2
-1
-1
-2
-3
-4
-5
1
2
3
𝑥
Math 110,
(1pts )
29/5/1431H
29. The curve 𝑓 (𝑥) =
Test 2–Version B
𝑥 − 2𝑥3 + 1
𝑥3 − 𝑥 + 5
11
has a horizontal asymptote at
(a)
 𝑦 = −2
(b) 𝑦 = 0
(c) 𝑦 = 5
(d) 𝑥 = −2
Solution:
To find the horizontal asymptote we take the limit as 𝑥 → ±∞ Note that both the
numerator and the denominator → ±∞, as 𝑥 → ±∞. To find this limit divided both the
numerator and the denominator by the highest power of 𝑥 in the denominator which is 𝑥3 .
So,
𝑥 − 2𝑥3 + 1
𝑥−
+1
𝑥3
= lim
lim
3
3
𝑥→±∞ 𝑥 − 𝑥 + 5
𝑥→±∞ 𝑥 − 𝑥 + 5
𝑥3
1
1
−2+ 3
2
𝑥
= lim 𝑥
1
5
𝑥→±∞
1− 2 + 3
𝑥
𝑥
0−2+0
= −2.
=
1−0+0
2𝑥3
Therefore 𝑦 = −2 is a horizontal asymptote.
(1pts )
∣𝑥 − 3∣ − 2
=
𝑥−1
(a) 1
30. lim
𝑥→1
(b) Does Not Exist
(c) 0
(d)
 −1
Solution:
A direct substation gives I.F. 0/0. Note that if 𝑥 > 1 or 𝑥 < 1 near(close) to 1 then
𝑥 − 3 < 0 ⇒ ∣𝑥 − 3∣ = −(𝑥 − 3).
∣𝑥 − 3∣ − 2
−(𝑥 − 3) − 2
= lim
𝑥→1
𝑥→0
𝑥−1
𝑥−1
−𝑥 + 3 − 2
= lim
𝑥→1
𝑥−1
−𝑥 + 1
= lim
𝑥→1 𝑥 − 1
−(𝑥 − 1)
= lim
𝑥→1 𝑥 − 1
= lim (−1)
lim
𝑥→1
= −1