Math 1910
TEST 1 Review Solutions
Page 1 of 4
1. Evaluate the limits in parts a through g analytically.
x+ x 4+ 4
6
x 2 − 5x − 6
(x − 6)(x + 1)
x +1 7
a. lim
=
=
b. lim
= lim
= lim
=
2
x →4 3x + 1
x →6
x →6 (x − 6)(x + 6)
x →6 x + 6
3(4) + 1 13
x − 36
12
x−5 x
x−5 x x+5 x
x 2 − 25x
x ( x − 25)
= lim
= lim
= lim
=
x →25 3x − 75
x →25 3x − 75 x + 5 x
x → 2 5 (3x − 75) x + 5 x
(
) x →25 3(x − 25)( x + 5 x )
x
25
25 1
lim
=
=
=
x →25 3 x + 5 x
(
) 3 25 + 5 25 150 6
c. lim
(
)
4 + sin x
4 + sin0
=
= −4
x →0 cos(x + π )
cos π
d. lim
f. lim −
x →−1
x+2
x +1
[ x] + 2
x→1
x →−2
x →−2
( x + 2) →1 and ( x + 1) → 0 , so there is an asymptote at x = −1.
x < −1, so x + 1 < 0. Then
g. lim
e. lim + f (x) = lim + (5x + 1) = −9
 +
x+2
x+2
= −∞.
must be negative   , so lim −
 −
x →−1 x + 1
x +1
0+2
= 1.
x +1
x +1
1 +1
[ x] + 2
[ x] + 2 1 + 2 3
[ x] + 2
lim+
=
= . ∴lim
: x > 1, so [ x ] → 1. ∴ lim−
does not exist.
x →1
x→1
x→1 x + 1
x +1
x +1
1 +1 2
lim
x +1
x →1 −
[ x] + 2
: x < 1, so [ x ] → 0 . ∴ lim−
[ x] + 2
x→1
=
x −1
x −1
x −1
lim−
= −1.
: x < 1, so x − 1 < 0 and x − 1 = −(x − 1) . Then lim−
x→1 x − 1
x →1 x − 1
x →1 x − 1
x −1
x −1
x −1
lim+
= 1. ∴lim
: x > 1, so x − 1 > 0 and x − 1 = (x − 1). Then lim+
does not exist.
x →1 x − 1
x →1 x − 1
x→1 x − 1
h. lim
sin (11x ) 1
sin(11x ) 1
11sin (11x ) 11
sin(11x ) 11
= lim
= lim
= lim
= .
x →0
13x
13 x →0
x
13 x→ 0
11x
13 x →0 11x
13
i. lim
j.
lim+
x →−6
k. lim−
x →3
x+6
x+6
x > −6, so x + 6 > 0 and x + 6 = x + 6. Then lim+
x →−6
[ x]
x < 3, so [ x ] → 2. ∴ lim−
x −1
x→ 3
[ x]
x −1
=
x+6
x+6
= lim+
= 1.
x + 6 x→ −6 x + 6
2
= 1.
3 −1
2
l. lim+ f (x) = lim (6x − 3) = 15. lim− f (x) = lim (x + 2) = 11. ∴lim f (x) does not exist.
x →3
x →3
x →3
x →3
x →3
( x − 1)2
x 2 − 2x + 1
2
= lim
m. lim
. ( x − 1) → 4 and ( x + 1) → 0 , so there is an asymptote at x = −1.
x →−1
x → −1 x + 1
x +1
 +
( x − 1) 2
( x − 1)2
= −∞.
Limit from the left: x < −1, so x + 1 < 0. Then
must be negative   , so lim −
 −
x →−1
x +1
x+1
 +
( x − 1) 2
( x − 1)2
= +∞.
Limit from the right: x > −1, so x + 1 > 0. Then
must be positive   , so lim +
 +
x →−1
x +1
x+1
Math 1910
Since lim −
x →−1
TEST 1 Review Solutions
( x − 1)
2
x+1
≠ lim+
x → −1
( x − 1)
x +1
2
, it follows that lim
( x − 1)
x +1
x →−1
Page 2 of 4
2
does not exist.
2. Evaluate the limits in parts a and b graphically.
a. lim f (x) = −2
b. lim f (x) does not exist
x →2
x →2
3. Determine the values of x for which the function is discontinuous and label each discontinuity as either
removable or nonremovable.
x 2 − 4x
x(x − 4)
f (x) =
=
2
55x − x x(55x − 1)
1
1
x = 0 is a removable discontinuity. 55x − 1 = 0 ⇒ x =
. x=
is a nonremovable discontinuity.
55
55
4. Determine the value of a that makes each function continuous for all values of x.
ax + 1 x > 2
a. f (x) =  2
x≤2
 x
2
Since ax + 1 and x are both polynomials, the only possible discontinuity is at x = 2 .
3
2
As x → 2, ax + 1→ 2a + 1 and x → 4 . Since the one-sided limits must be equal, a =
2
3x + 7
b. g(x) = 
5x − 1
x > a

x ≤ a
3a + 7 = 5a − 1⇒ a = 4
5. Evaluate the following limit and write an "epsilon-delta" proof. lim ( 5x + 2 ) = 17
x→3
ε
Proof. Let ε > 0 be arbitrary and let δ = . Then if x is chosen so that 0 < x − 3 < δ , we have
5
ε
x−3 <
5
5x −3 <ε
5x − 15 < ε
(5x + 2 ) − 17 < ε .
6. Find f ′(x) using the limit process.
a.
f (x + Δx) − f (x)
x + Δx − x
x + Δx − x x + Δx + x
f ′(x) = lim
= lim
= lim
Δx →0
Δx →0
Δx → 0
Δx
Δx
Δx
x + Δx + x
x + Δx − x
Δx
1
1
1
= lim
= lim
= lim
=
=
Δx →0 Δx
Δ
x→
0
Δx
→
0
x + Δx + x
Δx x + Δx + x
x + Δx + x
( x + x) 2 x
(
)
(
)
(
)
Math 1910
TEST 1 Review Solutions
Page 3 of 4
b.
(x + Δx)2 − 2(x + Δx) − (x 2 − 2x )
f (x + Δx) − f (x)
= lim
Δx →0
Δx →0
Δx
Δx
2
2
2
x + 2xΔx + ( Δx ) − 2x − 2Δx − x + 2x
2xΔx + (Δx )2 − 2Δx
Δx( 2x + Δx − 2)
= lim
= lim
= lim
Δx →0
Δx →0
Δx →0
Δx
Δx
Δx
= 2x − 2
f ′(x) = lim
7. Find all vertical asymptotes of each function:
x 2 + 5x
x(x + 5)
x
a. f (x) = 2
=
(x = 5)
b. f (x) =
(x = k π , where k is any integer but 0.)
x − 25 (x − 5)(x + 5)
sin x
8. Use the rules discussed in class to differentiate the following functions:
a. f (x) = 2x 5 − 8x + 3
b. g(x) = 63 x = 6x1 / 3
f ′(x) = 10x 4 − 8
g ′(x) = 2x −2 / 3 =
3
3
3 −5
x
5 =
5 =
(2x ) 32x 32
15
15
y ′ = − x −6 = −
32
32x 6
d. y =
2
x2 / 3
5
5 −1 0
x
10 =
7x
7
50
50
h ′(x) = − x −1 1 = − 1 1
7
7x
c. h(x) =
e. y = 3sin x
f. y = π 2 − 5cos x
y ′ = 3cos x
y ′ = 5sin x
9. Find all values of x for which each of the functions in parts a through d is not differentiable:
 3x + 4
x >0
a. f (x) = 
Only possible discontinuity is at x = 0.
4 + 3sin x x ≤ 0
d
d
(3x + 4) = 3 and ( 4 + 3sin x ) = 3cos x . Both of these derivatives exist for all values
lim f (x) = 4.
x →0
dx
dx
of x. As x → 0 , 3 → 3 and 3cos x → 3cos(0) = 3. ∴ f is differentiable for all values of x
b. f (x) = x 4 / 3 f ′(x) =
c. x = 2 (cusp)
4 1/ 3 4 3
x =
x , which is defined for all values of x.
3
3
d. x = 2 (discontinuity)
10. Let f (x) = x 3 − 3x + 1.
a. Find an equation of the tangent line at (5,111).
f ′(x) = 3x 2 − 3 m = f ′ (5) = 72
y − 111 = 72(x − 5) or y = 72x − 249
b. For which values of x does the graph of f have a horizontal tangent line?
The values of x for which f ′(x) = 3x 2 − 3 = 0 : x = ±1
c. For which values of x does the graph of f have a tangent line parallel to graph of y = 7x + 1?
10
2
The values of x for which f ′(x) = 3x − 3 = 7: x = ±
.
3
Math 1910
TEST 1 Review Solutions
Page 4 of 4
11.
12. Consider the function f (x) = x 2 + x .
a. Find the average rate of change in f over the interval [-1,2].
f (2) − f (−1)
6−0
=
=2
2 − (−1)
2 − (−1)
b. Find the instantaneous rate of change in f at x = 2.
f ′(x) = 2x + 1
f ′(2) = 5
13. An object dropped off the top of an 800-foot building is s feet above the ground t seconds after being
released, where s(t) = −16t 2 + 800.
a. Find the velocity function. v(t) = s′ (t) = −32t
b. Find the velocity of the object 2 seconds after release. v(2) = −32(2) = −64 ft/sec
c. How long does it take the object to hit the ground? Set s(t) = 0 and solve for t.
−16t 2 + 800 = 0 ⇒16t 2 = 800 ⇒ t 2 = 50 ⇒ t = 50 seconds
d. What is the velocity of the object as it strikes the ground?
e. What is the average velocity of the object between t = 1 and t = 3?
s(3) − s(1) 656 − 784
=
= −64 ft/s
3 −1
2
v( 50) = −32 50 ≈ −226.27 ft/sec