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WorksheetsUnit 1: Limits and Continiutiy
Total questions: 143
Worksheet time: 5hrs 16mins
0
2
3
4
Find x→2− limf(x)
-1
5
0
DNE
Find x→2+ limf(x)
-1
5
0
DNE
Find x→2 limf(x)
-1
5
0
DNE
Find f(2)
-1
5
0
DNE
Find x→−1− limf(x)
4
0
-1
DNE
Find x→−1+ limf(x)
4
0
-1
DNE
Find x→−1 limf(x)
4
0
-1
DNE
Find f(−1)
4
0
-1
DNE
Find x→−4− limf(x)
-2
3
-4
DNE
Find x→−4+ limf(x)
-2
3
-4
DNE
Find x→−4 limf(x)
-2
3
-4
DNE
Find f(−4)
-2
3
-4
DNE
Find f(4)
2
4
-4
DNE
Find x→4 limf(x)
2
0
-4
DNE
Find x→4+ limf(x)
-2
2
-4
DNE
If x→climf(x)=3 , x→climg(x)=−2 , and x→climh(x)=4 then find x→clim3h(x)−2g(x) .
16
4
8
12
If x→climf(x)=3 , x→climg(x)=−2 , x→climh(x)=4 then find x→clim[f(x)⋅5g(x)]
-30
-40
60
-80
If x→climf(x)=3 , x→climg(x)=−2 , x→climh(x)=4 then find x→clim[7−g(x)]2
81
25
9
45
If x→climf(x)=3 , x→climg(x)=−2 , x→climh(x)=4 then find x→clim(h(x)−g(x)2f(x)+3h(x))
3
2
1
0
If x→climf(x)=3 , x→climg(x)=−2 , x→climh(x)=4 then find x→clim[h(x)⋅(f(x)+6)]
36
16
-18
-20
If x→climf(x)=3 , x→climg(x)=−2 , x→climh(x)=4 then find x→clim[4−g(x)f(x)2 ]
3/2
8/3
2/3
9/2
Use the graph above to solve: x→−1limg(f(x))
-7
-2
-6
DNE
Use the graph above to solve: x→−3limg(f(x))
-7
-2
-6
DNE
x → 12lim(3 x−7)=L
Submit L
(a)
x → 19lim(2 x 2−36 x−5)=L
Submit L
(a)
x → − 6lim(3 x 3+19 x 2+4 x−20)=L
Submit L
(a)
x → 8lim( x+8x 3+512 )=L
Submit L
(a)
Which of the following limits cannot be found using direct substitution?
( Check all that apply )
x → 9lim ( 2 x 3−15 x 2 −28 x +16 )
x→ 17lim (x−17x 2−289)
x → 5lim ( x 2+3 x−10x 2−4 x−45 )
x → 4lim ( x 2−7 x+12x 2+3 x−28 )
x → 7lim 5 x+1
Polynomial function P(x)
x → clim P(x)=P(c)
Always
Sometimes
Never
x → 26lim13 x−14=L
Submit L
(a)
x → 6πlimcos(x)=BA
Submit A,B
(a)
Rational function R(x)
x → clim R(x)=R(c)
Always
Sometimes
Never
Which of the following limits cannot be found using direct substitution?
( Check all that apply )
x → 5lim ( x−5x+41−91 )
x→ 6lim (x+6x 3+216)
x → 11lim ( x−11x+12−23 )
x → 4lim ( x−4x 5−1024 )
x → 33lim ( x−20x−8−41 )
x→−3lim(2x2+x−5)=
If the limit does not exist, write "DNE."
(a)
x→5lim3x−11=
If the limit does not exist, write "DNE."
(a)
x→0lim(−5)=
If the limit does not exist, write "DNE."
(a)
x→−1lim xx2−3x+9=
If the limit does not exist, write "DNE."
(a)
x→2lim x−2x2+3x=
If the limit does not exist, write "DNE."
(a)
x→−3lim x+32x2+7x+3=
If the limit does not exist, write "DNE."
(a)
x→7lim x−7−x2+13x−42=
If the limit does not exist, write "DNE."
(a)
x→0lim 3xsin(4x)=
If the limit does not exist, write "DNE."
(a)
x→0lim 2x1−cos(4x)=
If the limit does not exist, write "DNE."
(a)
x→0lim 4x2sin2(5x)=
If the limit does not exist, write "DNE."
(a)
If f is the function defined by f(x)=x2+2x−15x2−9, then x→3limf(x) is
0
159
43
nonexistent
x→3lim x3−9xx−3 is
0
181
1
nonexistent
It is known that x→0lim 2xsin(2x)=1. What is x→0lim 8xcot(2x)cos(5x)?
0
81
41
nonexistent
It is known that x→0lim 2xsin(2x)=1. What is x→0lim 6xsec(3x)tan(2x)?
0
61
31
nonexistent
If f is the function defined by f(x)=x2+x−6x2−4, then x→2limf(x) is
0
32
54
nonexistent
x→−4lim 17 = ....
17
−68
−4
−17
0
x→−5lim 2x+5 = ....
0
5
∞
25
52
f(x) = 5x−14x−5
What law is appropriate to use to get the limit of the function above?
Quotient law
Sum law
Root law
Identity function law
x→ylim 5(4x2)
What law is appropriate to use to get the limit of the function above?
Constant Multiple Law
Sum law
Root law
Identity function law
Find the t→2lim t2−5t
-6
-10
8
14
z→−2lim z3+2
What law is appropriate to use to get the limit of the function above?
Constant Multiple Law
Sum law
Root law
Identity function law
Find the x→2lim f(x)×g(x) if f(x) = x2+3; g(x)= 2
2
14
11
12
x→3lim (3x)4
What law is appropriate to use to get the limit of the function above?
Difference Law
Sum law
Root law
Power law
x→3lim 3x+4
What law is appropriate to use to get the limit of the function above?
Identity Law
Difference Law
Root law
Quotient Law
If f(x) = 4x and g(x)=2x , find x→−2limf(x) − g(x) .
-4
8
16
0
x→25lim x−25x−5
2
−31
101
178
x→16lim x−4x−16
14
13
1
8
x→2lim x−2x−1−1
−41
−89
−57
21
x→2lim −x2−5x+6x−2
−5
−6
1
−3
x→5lim x−5x2−4x−5
−2
5
15
6
x+1x2−1 x→−1lim
−2
−11
0
6
x→−3lim −x+3x2−9
13
12
6
−2
Identify the type of location of the discontinuity/discontinuities.
Infinite at x = 4
The function is continuous
Removable at x = 4
Jump at x = 4
Where is the infinite discontinuity (Vertical Asymptote)?
x= -5
x= 5
x= 6
x= -6
Where is the removable discontinuity (hole)?
x= 4
x= -4
x= 5
x= -5
Evaluate the limit: x→0lim x5x2−20x
1/20
20
-20
-1/20
Find x→2− limf(x)
-1
5
0
DNE
Find x→2+ limf(x)
-1
5
0
DNE
Find x→2 limf(x)
-1
5
0
DNE
Find f(2)
-1
5
0
undefined
Find x→−1− limf(x)
4
0
-1
DNE
Find x→−1+ limf(x)
4
0
-1
DNE
Find x→−1 limf(x)
4
0
-1
DNE
Find x→−4− limf(x)
-2
3
-4
DNE
Find x→−4+ limf(x)
-2
3
-4
DNE
Find x→−4 limf(x)
-2
3
-4
DNE
Find f(−4)
-2
3
-4
undefined
Find f(4)
2
4
-4
undefined
Find x→4 limf(x)
2
0
-4
DNE
The graph of the function, f(x) is shown in the graph. What is x→−1limf(x)?
1
2
5
dne
The graph of the function, f(x) is shown in the graph. What is x→2limf(x)?
1
2
5
dne
The graph of the function, f(x) is shown in the graph. What is x→2+limf(x)?
1
2
5
dne
The graph of the function, f(x) is shown. What is x→2+limf(x)?
1
3
4
dne
The graph of the function, f(x) is shown. What is x→2−limf(x)?
1
3
4
dne
The graph of the function, f(x) is shown. What is f(2)?
1
3
4
dne
The graph of the function, f(x) is shown above. What is x→2+limf(x)?
3
2
1
dne
4
The graph of the function, f(x) is shown above. What is x→2−limf(x)?
3
2
1
dne
4
The graph of the function, f(x) is shown above. What is x→2limf(x)?
3
2
1
dne
4
Determine the type(s) and location(s) of all discontinuitie(s)
jump at x = 1
removable and x = 1
infinite at x = 1
jump at x = -3
y=-x(x+2)(x-1)
-4
-1
0
DNE
0
1
2
DNE
2
4
32
DNE
Let f be a continuous function for which f(-2)=1 and f(5)=-3. The Intermediate Value Theorem guarantees that
f(c)=2 for at least one c between -3 and 1
f(c)=0 for at least one c between -2 and 5
f(c)=0 for at least one c between -3 and 1
f(c)=2 for at least one c between -2 and 5
I only
II only
I, II, III
None
f is continuous for all real numbers x
The domain of f ix (-∞,3)(3,∞)
what type of discontinuity is represented in the graph
continuous
finite jump discontinuity
infinite discontinuity
removable discontinuity
if
1.) f(a) is defined or undefined
2.) limit of f(x) as x approaches to a - is not equal to f(x) as x approaches to a+
then what kind of discontinuity is this?
finite jump discontinuity
infinite discontinuity
removable discontinuity
f (a) = x→alimf (x) what kind of discontinuity if
f (a) is undefined
the limit exists
and
continuous
finite jump discontinuity
infinite discontinuity
removable discontinuity
x→0lim x(x+9) − 9
what is the limit?
1
2
0
-1
x→−∞lim(8x5−6x4+20x −113x5+4x3−15x2+7)
(a)
x→∞lim(13x+112x3+17x−1)
∞
−∞
35
0
x→−∞lim(x5−35x)
(a)
x→−∞lim(−3x+105x2−17)
∞
−∞
35
0
x→∞lim(5x4−3x3−11x2 +71x4−11x2+7)
(a)
x→−∞lim(−5+x34)
(a)
x→−7+lim(x+75)
∞
−∞
DNE
75
x→7−lim(x2−4x−21x2−9)
∞
−∞
DNE
1
x→−5lim(x2+4x−5x−1)
∞
−∞
DNE
1
x→−3lim((x+3)2−4)
∞
−∞
DNE
1
If a function f has a domain
(−∞, ∞) , then it is _______________.increasing
extraneous
continuous
undefined
For the function
f(x)=⌊x⌋ , determine an interval where f is continuous.[0.5, 1.5]
[-1, 1]
[-2, 0]
[1, 1.5]
Suppose t is a continuous function containing the ordered pairs shown in the table. On which interval(s) must t attain a value of -2.5? Select all that apply.
(1, 2)
(2, 3)
(3, 4)
(4, 5)
(5, 6)
Suppose f is a continuous function containing the ordered pairs shown in the table. On which interval(s) must f attain a value of 5.9? Select all that apply.
(-3, -2)
(-2, -1)
(0, 1)
(1, 2)
(2, 3)
Suppose h is a continuous function containing the ordered pairs shown in the table. On which interval must h have a zero?
(12.163, 12.164)
(12.164, 12.165)
(12.165, 12.166)
(12.166, 12.168)
Assume f(x) is continuous over the interval [-2, 3]. The function f has at least how many zeros?
1
2
3
4
Determine the interval(s) for which the function is continuous. Select all that apply.
(p, 0)
[0, q]
[q, r]
(r, s)
(q, s)
Use the Intermediate Value Theorem to find an interval between two consecutive integers that contains a zero of the function
h(x)=2x−x2−lnx[-1, 0]
[0, 1]
[1, 2]
(-1, 1)
Consider the function
f(x)=x−31−5 . Can you conclude that there must be a zero between f(2) and f(3.1)?Yes, because f(2) is negative and f(3.1) is positive.
No, because f(2) is negative and f(3.1) is also negative.
Yes, because f(2) is positive and f(3.1) is negative.
No, because there is a discontinuity at x = 3.
No, because f(2) is positive and f(3.1) is also positive.
A hot apple pie is left on a window sill to cool. Suppose its temperature (degrees F) after x minutes is given by
f(x)=90e−0.61x+70 . Give an interval in which its temperature will first be under 100°F(0, 0.5)
(0.5, 1)
(1, 1.5)
(1.5, 2)
