YORK UNIVERSITY
Faculty of Science and Engineering
Faculty of Liberal Arts and Professional Studies
MATH 1013 3.00 B
Test #2
Solutions
1. (6 pts) Use the − δ definition of the limit to prove that
2x2 − 11x + 5
= 9.
x→5
x−5
lim
Answer:
Guessing a value for δ.
2x2 − 11x + 5
=9
x→5
x−5
lim
0 < |x − 5| < δ =⇒ |
if (∀ > 0) (∃δ() > 0) :
2x2 − 11x + 5
− 9| < .
x−5
Now,
|
2x2 − 11x + 5
(2x − 1)(x − 5)
− 9| < ⇐⇒ |
− 9| < x−5
x−5
⇐⇒ |(2x − 1) − 9| < ⇐⇒ |2x − 10| < ⇐⇒ |2(x − 5)| < ⇐⇒ |x − 5| < .
2
Hence, we should choose δ = .
2
Showing that this δ works.
Given > 0, choose δ = . If 0 < |x − 5| < δ, then
2
|
2x2 − 11x + 5
(2x − 1)(x − 5)
− 9| = |
− 9|
x−5
x−5
= |(2x − 1) − 9| = |2x − 10| = |2(x − 5)| < 2δ = 2 = .
2
Note, that any δ ∈ (0, ] will also satisfy the definition.
2
Continues...
Name:
Student No:
2. (3 + 3 + 3 pts) Evaluate each of the following limits:
(3x + 2)2
;
x→∞
2x2
Answer:
(a) lim
(3x + 2)2
9x2 + 12x + 4
lim
=
lim
= lim
x→∞
x→∞
x→∞
2x2
2x2
9x2 +12x+4
x2
2x2
x2
= lim
9+
x→∞
12
x
+
4
x2
2
1
12
4
9
9
= ( lim 9 + lim
+ lim 2 ) = (1 + 0 + 0) = .
x→∞ x
x→∞ x
2 x→∞
2
2
tan 2θ
;
θ→0 sin 5θ
Answer:
(b) lim
tan 2θ
sin 2θ 1
2 sin 2θ 1
5θ
= lim (
) = lim (
)
θ→0 sin 5θ
θ→0 cos 2θ sin 5θ
θ→0
2θ cos 2θ 5 sin 5θ
2
sin 2θ
1
5θ
= (lim (
)(lim
)(lim
)
θ→0 cos 2θ θ→0 sin 5θ
5 θ→0 2θ
2
sin 2θ
1
1
2
1 1
2
= (lim
)(
)(
)= ·1· · = ,
sin
5θ
5 θ→0 2θ
limθ→0 cos 2θ limθ→0 5θ
5
1 1
5
lim
∵ θ → 0 =⇒ 2θ → 0 & 5θ → 0.
(c) lim arctan(
x→2
Answer:
x2 − 4
).
3x2 − 6x
lim arctan(
x→2
x2 − 4
(x − 2)(x + 2)
x+2
) = lim arctan(
) = lim arctan(
)
2
x→2
x→2
3x − 6x
3x(x − 2)
3x
= arctan lim (
x→2
2+2
2
x+2
) = arctan(
) == arctan( ).
3x
3·2
3
Continues...
2
Name:
Student No:
3. (3 + 2 + 2 pts) Let f (x) =
p
1 − |x − 1|.
(a) Determine expression for f 0 (x).
Answer:
The domain of f (x) is all the values of x, such that 1 − |x − 1| ≥ 0 or |x − 1| ≤ 1 or
−1 ≤ x − 1 ≤ 1 or 0 ≤ x ≤ 2.
Since
x − 1, if x ≥ 1
|x − 1| =
1 − x, if x < 1,
p
√
1 − (1 − x) = x,
if 0 ≤ x ≤ 1
p
√
f (x) =
1 − (x − 1) = 2 − x, if 1 < x ≤ 2.
And since
lim
x→1−
f (x) − f (1)
1
1
f (x) − f (1)
= 6= − = lim
,
x−1
2
2 x→1+
x−1

1

if 0 < x < 1
 2√x ,
0
DNE,
if x = 1
f (x) =

 − √1 , if 1 < x ≤ 2.
2 2−x
(b) For what value(s) of x (if any) is f (x) not differentiable? Explain.
Answer:
¿From part (a), f (x) is not differentiable at x = 0, 2 (end points) and at x = 1, where
f 0 (x) does not exist.
(c) Is f (x) continuous at every point x of its domain? Explain.
Answer: √
√
x and 2 − x are continuous functions.
Also,
lim f (x) = 1 = lim f (x)
x→1−
x→1+
and f (1) = 1, implies that f (x) is continuous at x = 1.
Therefore, f (x) continuous at every point x of its domain.
Continues...
3
Name:
Student No:
4. (6 + 4 pts) Consider the function g(x) = cos(2x).
(a) Use the definition of the derivative to show that g 0 (x) = −2sin(2x).
Note: No points will be given to any other solutions.
Answer:
g(x + h) − g(x)
cos[2(x + h)] − cos 2x
= lim
h→0
h→0
h
h
cos(2x) cos(2h) − sin(2x) sin(2h) − cos(2x)
cos(2x + 2h) − cos 2x
= lim
= lim
h→0
h→0
h
h
cos(2h) − 1
sin 2h
cos 2h − 1
sin 2h
= lim [cos(2x)
− sin(2x)
] = lim [2 cos(2x)
− 2 sin(2x)
]
h→0
h→0
h
h
2h
2h
cos(2h) − 1
sin 2h
= 2 cos(2x) lim
−2 sin(2x) lim
= 2 cos(2x)·0−2 sin(2x)·1 = −2 sin(2x),
h→0
h→0 2h
2h
∵ h → 0 =⇒ 2h → 0.
g 0 (x) = lim
(b) Use the formula from part (a) to determine a slope-intercept equation of the tangent
π 1
line and the normal line to the graph of y = cos(2x) at the point ( , ).
6 2
Answer:
π 1
The slope of the tangent line to the graph of y = cos(2x) at the point ( , ) is equal to
6 2
√
√
π
π
π
3
y 0 |x= π6 = g 0 ( ) = −2 sin(2 · ) = −2 sin( ) = −2 ·
= − 3.
6
6
3
2
Hence, a point-slope equation of the tangent line to the graph of y = cos(2x) at the
π 1
point ( , ) is
6 2
√
1
π
y − = − 3(x − ),
2
6
and a slope-intercept equation of the tangent line will be
√
√
3π
1
y = − 3x + ( +
).
2
6
π 1
The slope of the normal line to the graph of y = cos(2x) at the point ( , ) is equal to
6 2
1
1
− √ = √ . Thus, a point-slope equation of the normal line to the graph of y = cos(2x)
− 3
3
π 1
at the point ( , ) is
6 2
1
1
π
y − = √ (x − ),
2
6
3
and a slope-intercept equation of the normal line will be
π
1
1
y = √ x + ( − √ ).
2
3
6 3
Continues...
4
Name:
Student No:
5. (3 + 3 + 3 pts)
Hint: You may need to simplify a function before differentiating.
√
√
3
x2 + x − 1
dy
(a) If y =
, then
=
x
dx
Answer:
√
√
2
1
3
1
1
x2 + x − 1
x3 + x2 − 1
y=
=
= x− 3 + x− 2 − x−1 .
x
x
So,
4
3
1
dy
d −1
1
1
=
(x 3 + x− 2 − x−1 ) = (− )x− 3 + (− )x− 2 − (−1)x−2
dx
dx
3
2
3
4
1
1
= x−2 − x− 2 − x− 3 .
2
3
x2 − 1
,
x2 + 1
Answer:
(b) If y =
then
dy
=
dx
d
d
(x2 + 1) dx
(x2 − 1) − (x2 − 1) dx
(x2 + 1)
dy
d x2 − 1
=
( 2
)=
dx
dx x + 1
(x2 + 1)2
=
2x(x2 + 1) − 2x(x2 − 1)
2x3 + 2x − 2x3 + 2x
4x
=
= 2
.
2
2
2
2
(x + 1)
(x + 1)
(x + 1)2
sin x + cos x
,
tan x
Answer:
(c) If y =
y=
then
dy
=
dx
sin x
cos x
cos2 x
1 − sin2 x
+
= cos x +
= cos x +
tan x tan x
sin x
sin x
1
= cos x +
− sin x = cos x + csc x − sin x.
sin x
So,
d
dy
=
(cos x + csc x − sin x) = − sin x − csc x cot x − cos x
dx
dx
= −(sin x + csc x cot x + cos x).
Continues...
5
Name:
Student No:
6. (2 + 2 + 2 pts) If a rock thrown upward on the planet Mars with a velocity of 10 m/s, its
height (in meters) after t seconds is given by H = 10t − 1.86t2 .
(a) Find the velocity of the rock when t = a.
Answer:
v(a) =
dH
|t=a .
dt
dH
d
= (10t − 1.86t2 ) = 10 − 3.72t.
dt
dt
So, v(a) = 10 − 3.72a.
(b) When the rock will hit the surface?
Answer:
The rock will hit the surface when
H = 0 ⇐⇒ 10t − 1.86t2 = 0 ⇐⇒ t(10 − 1.86t) = 0 ⇐⇒ t = 0 or 1.86t = 10.
10
(s).
1.86
(c) With what velocity will the rock hit the surface?
Answer:
The velocity of the rock when it hits the surface is
Hence, the rock will hit the surface when t =
v(
10
10
m
) = 10 − 3.72
= 10 − 20 = −10 ( ).
1.86
1.86
s
Continues...
6
Name:
Student No:
7. (12 pts) For each statement indicate whether it is always TRUE or sometimes FALSE.
Note: For this question, each correct answer is worth 1.5 point and each incorrect answer
is worth −0.5 (negative half!) point. If the number of incorrect answers is more than three
times greater than the number of correct ones, then the total mark will be zero. If you don’t
know the answer, don’t write anything. For this question only, you do NOT need to explain
your answer or show your work.
Statement
TRUE/FALSE
If lim f (x) = lim f (x), then f (x) is continuous at x = a.
FALSE
If a function f (x) is continuous, then the limit lim f (x) exists
TRUE
x→a−
x→a+
x→a
for every a in the domain of f (x).
If f (x) is continuous on [−1, 1], and f (−1) = 2 and f (1) = 3, then there
exists a number c, such that |c| < 1 and f (c) = e.
TRUE
If g(x) is a trigonometric function, then lim g(x) = g(b), for every real
FALSE
x→b
number b.
If (f + g)(x) is differentiable, then the both f (x) and g(x) are differentiable.
FALSE
A function may have two different horizontal asymptotes.
TRUE
If g 0 (a) exists, then lim g(x) = g(a).
TRUE
d2 y
dy
= ( )2 .
2
dx
dx
FALSE
x→a
THE END.
7