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2.1-2.2 Review – Calculus
Honors Precalculus – Meinke
Name: ______________________________
No Calculator
Definition: 𝑓(π‘₯) = 𝑖𝑛𝑑π‘₯ is called the β€œgreatest integer function”. It is also called the β€œrounding down
function”.
Examples: 𝑖𝑛𝑑2 = _____
5
𝑖𝑛𝑑 2 = _____
5
1
𝑖𝑛𝑑(βˆ’ 2) = _____
𝑖𝑛𝑑 2 = _____
Evaluate the following limits.
Direct substitution:
1.) lim 2
π‘₯β†’3
4.) lim
3.) lim 3βˆ’2π‘₯
5.) lim 𝑖𝑛𝑑(βˆ’π‘₯ βˆ’ 0.5)
6.) lim [𝑖𝑛𝑑(π‘₯ + 1) βˆ’ 𝑖𝑛𝑑π‘₯ 2 ]
π‘₯β†’5βˆ’
π‘₯ 2 +4
π‘₯β†’βˆ’2 π‘₯βˆ’2
6π‘₯βˆ’1
2.) lim πœ‹π‘₯
π‘₯β†’βˆ’3
π‘₯β†’0
π‘₯β†’3βˆ’
Algebraic Simplification, Then Substitution:
7.) lim
π‘₯ 2 βˆ’4
8.) lim
π‘₯β†’2 π‘₯βˆ’2
10.) lim
π‘₯β†’βˆ’π‘Ž
2π‘₯ 2 βˆ’π‘₯βˆ’15
π‘₯β†’3
π‘₯ 2 +2π‘Žπ‘₯+π‘Ž2
π‘₯+π‘Ž
π‘₯βˆ’3
π‘₯ 2 βˆ’2π‘₯
11.) lim π‘₯ 2 +π‘₯βˆ’6
π‘₯β†’2
9.) lim
π‘₯+7
π‘₯β†’βˆ’7+ π‘₯ 2 βˆ’49
12.) lim
𝑠𝑖𝑛5π‘₯
π‘₯β†’0 2π‘₯
Limits at Infinity:
βˆ’5
13.) lim
π‘₯β†’βˆž
14.) lim
π‘₯2
π‘₯β†’βˆž
15.) lim
π‘₯β†’βˆ’βˆž 5
6π‘₯ 3 +4π‘₯
16.) lim
π‘₯2
βˆ’π‘₯ 3 βˆ’4π‘₯ 2 +5
19.) lim π‘‘π‘Žπ‘›π‘₯
π‘₯β†’βˆ’βˆž
π‘₯β†’βˆž
π‘₯ 5 βˆ’1
17.) lim
π‘₯β†’βˆ’βˆž 66π‘₯ 2 +π‘₯βˆ’1
20.) lim
π‘₯β†’βˆž
1
1+𝑒 βˆ’π‘₯
2π‘₯ 3 +π‘₯ 2 +2
π‘₯ 4 βˆ’75
18.) lim 𝑠𝑖𝑛π‘₯
π‘₯β†’βˆž
21.) lim
1
π‘₯β†’βˆ’βˆž 1+𝑒 βˆ’π‘₯
βˆ’1, π‘₯ < 0
Definition: 𝑠𝑔𝑛π‘₯ = { 0, π‘₯ = 0
1, π‘₯ > 0
22.) lim
π‘₯2
π‘₯β†’βˆ’βˆž 𝑠𝑔𝑛π‘₯
23.) lim 2𝑠𝑔𝑛π‘₯
π‘₯β†’βˆž
24.) lim
π‘₯3
π‘₯β†’βˆ’βˆž 𝑠𝑔𝑛π‘₯
Limits at V.A:
25.) lim
π‘₯ 2 +4
π‘₯β†’2 π‘₯βˆ’2
π‘₯ 2 βˆ’4
28.) lim (π‘₯βˆ’2)3
π‘₯β†’2
π‘₯ 2 +4
26.) lim (π‘₯βˆ’2)2
π‘₯β†’2
4βˆ’π‘₯ 2
29.) lim (π‘₯βˆ’2)3
π‘₯β†’2
27.) lim
π‘₯ 2 +4
π‘₯β†’βˆ’2 π‘₯+2
π‘₯ 2 βˆ’4
30.) lim (π‘₯βˆ’2)2
π‘₯β†’2
1
1
31.) lim (π‘₯ βˆ’ π‘₯ 2 )
π‘₯β†’0+
βˆ’1
32.) lim
π‘₯β†’2βˆ’ 2βˆ’π‘₯
1
33.) lim (π‘₯βˆ’3)2
π‘₯β†’3
Other Limits:
34.) lim 𝑖𝑛𝑑π‘₯
35.) lim 𝑖𝑛𝑑π‘₯
π‘₯β†’4
37.) lim
π‘₯β†’4.5
|π‘₯|
38.) lim
π‘₯β†’0 π‘₯
|π‘₯|
36.) lim 𝑖𝑛𝑑π‘₯
π‘₯β†’βˆ’4
39.) lim
π‘₯β†’0βˆ’ π‘₯
|π‘₯|
π‘₯β†’0+ π‘₯
Definition: If lim 𝑓(π‘₯) = 𝐿 and/or lim 𝑓(π‘₯) = 𝑀, where L and M are finite, then 𝑓(π‘₯) has a H.A. at
π‘₯β†’βˆž
π‘₯β†’βˆ’βˆž
𝑦 = 𝐿 and/or 𝑦 = 𝑀.
Find the horizontal asymptote(s).
40.) 𝑦 =
1
π‘₯
1+π‘π‘œπ‘ ( )
1
1+
π‘₯
1
43.) 𝑦 = 1+𝑒 π‘₯
41.) 𝑦 =
44.) 𝑦 =
2π‘₯+𝑠𝑖𝑛π‘₯
π‘₯
βˆ’3π‘₯+1
2π‘₯+2
𝑠𝑖𝑛π‘₯
42.) 𝑦 = 2π‘₯ 2 +π‘₯
45.) 𝑦 = 𝑠𝑒𝑐π‘₯
46.) Let 𝑓(π‘₯) = {
βˆ’π‘₯ 2 βˆ’ 1, π‘₯ < 1
.
π‘₯ βˆ’ 1, π‘₯ β‰₯ 1
Find the following:
π‘Ž. ) lim 𝑓(π‘₯)
π‘₯β†’1βˆ’
b.) lim 𝑓(π‘₯)
c.) lim 𝑓(π‘₯)
π‘₯β†’1+
π‘₯β†’1
47.) Given the graph of g(x) below, find the following:
a.) lim 𝑔(π‘₯)
π‘₯β†’0
b.) lim 𝑔(π‘₯)
π‘₯β†’βˆž
c.) lim 𝑔(π‘₯)
π‘₯β†’βˆ’βˆž
d.) lim 𝑔(π‘₯)
π‘₯β†’1+
48.) Sketch a function p(x) with the following conditions:
a.) p(x) has a limit at x = -2.
b.) p(-2) is undefined.
c.) lim 𝑝(π‘₯) = 2
π‘₯β†’βˆž
d.) lim 𝑝(π‘₯) = 2
π‘₯β†’βˆ’βˆž
e.) lim 𝑝(π‘₯) = 1
π‘₯β†’1βˆ’
f.) lim 𝑝(π‘₯) = 3
π‘₯β†’1+
g.) p(1) = -1
h.) lim 𝑝(π‘₯) = βˆ’βˆž
π‘₯β†’βˆ’1
Find the right and left-end behavior models.
49.) 𝑓(π‘₯) = βˆ’π‘’ βˆ’π‘₯ + π‘₯ 8
50.) 𝑔(π‘₯) = π‘₯ + 𝑠𝑖𝑛π‘₯
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