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Summary AP Caculus: Limits and Continuity

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This is a summary of the concept of Limits and Continuity. This concept is crucial for learning advanced calculus, which includes topics such as differentiation and integration.

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  • March 5, 2021
  • March 5, 2021
  • 2
  • 2020/2021
  • Summary
  • Secondary school
  • 5
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Limits and Continuity
Vague Definition of Limits
Let 𝑎 𝜖 ℝ and f be a function defined on an open interval containing 𝑎 (except possibly at 𝑎). Suppose L is a real number such that f(x) is as
close to L as we please whenever x is sufficiently close (but not equal to 𝑎). The number L is the limit of f(x) as x approaches 𝑎 and can be
written as lim 𝑓(𝑥) = 𝐿.
𝑥→𝑎
Limit Laws
Let a, c, L and M be real numbers and suppose that lim 𝑓(𝑥) = 𝐿 and lim 𝑔(𝑥) = 𝑀. Then
𝑥→𝑎 𝑥→𝑐
lim 𝑐𝑓(𝑥) = 𝑐 lim 𝑓(𝑥) lim [ 𝑓(𝑥) ± 𝑔(𝑥)] = lim 𝑓(𝑥) ± lim 𝑔(𝑥)
𝑥→𝑎 𝑥→𝑎 𝑥→𝑎 𝑥→𝑎 𝑥→𝑎
(Scalar Multiplicity) (Linearity)
lim [ 𝑓(𝑥) × 𝑔(𝑥)] = lim 𝑓(𝑥) × lim 𝑔(𝑥) lim 𝑓(𝑥)
𝑓(𝑥) 𝑥→𝑎
𝑥→𝑎 𝑥→𝑎 𝑥→𝑎
lim = provided that 𝑀 ≠ 0
𝑥→𝑎 𝑔(𝑥) lim 𝑔(𝑥)
𝑥→𝑎


Squeeze Theorem
𝐿𝑒𝑡 𝑎 ∈ 𝑅 𝑎𝑛𝑑 𝑙𝑒𝑡 𝑙 𝑏𝑒 𝑎𝑛 𝑜𝑝𝑒𝑛 𝑖𝑛𝑡𝑒𝑟𝑣𝑎𝑙 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝑎 ∈ 𝑙.
𝐼𝑓 𝑔(𝑥) ≤ 𝑓(𝑥) ≤ ℎ(𝑥) 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 ∈ 𝑙 \{𝑎}, 𝑎𝑛𝑑 𝑙𝑖𝑚 𝑔(𝑥) = 𝑙𝑖𝑚 ℎ(𝑥) = 𝐿 , 𝑡ℎ𝑒𝑛 𝑙𝑖𝑚 𝑓(𝑥) = 𝐿 .
𝑥→𝑎 𝑥→𝑎 𝑥→𝑎
1
Example: Find 𝑙𝑖𝑚𝑥 2 cos ( 2 )
𝑥→0 𝑥
Solution.
1
−1 ≤ 𝑐𝑜𝑠 (𝑥2 ) ≤ 1
1
−𝑥 2 ≤ 𝑥 2 𝑐𝑜𝑠 ( 2 ) ≤ 𝑥 2
𝑥
1
𝑆𝑖𝑛𝑐𝑒 𝑙𝑖𝑚 (−𝑥 2 ) = 𝑙𝑖𝑚 𝑥 2 = 0, 𝑖𝑡 𝑓𝑜𝑙𝑙𝑜𝑤𝑠 𝑡ℎ𝑒 𝑆𝑞𝑢𝑒𝑒𝑧𝑒 𝑇ℎ𝑒𝑜𝑟𝑒𝑚 𝑡ℎ𝑎𝑡 𝑙𝑖𝑚𝑥 2 𝑐𝑜𝑠 ( 2 ) = 0
𝑥→0 𝑥→0 𝑥→0 𝑥
sin 𝑥 sin 𝑛𝑥
Important Limits: 𝑙𝑖𝑚 = 1 and 𝑙𝑖𝑚 =𝑛
𝑥→0 𝑥 𝑥→0 𝑥
1−cos x 1−cos3 x
Example 1: Find lim Example 2: Find lim
x→0 x2 x→0 x2
Solution. Solution.
1 − cos x 𝐿𝑒𝑡 𝑓(𝑥) = 1 − 𝑐𝑜𝑠 3 𝑥
lim
x→0 x2 𝑓 ′ (𝑥) = 3𝑐𝑜𝑠 2 𝑥 𝑠𝑖𝑛𝑥
(1 − cos x)(1 + cosx)
= lim 𝑓 ′′ (𝑥) = 3(𝑐𝑜𝑠𝑥(1 − 3𝑠𝑖𝑛2 𝑥))
x→0 x 2 (1 + cosx)
𝑓 ′′ (0)𝑥 2 3
12 − cos2 x 𝑓(𝑥) = 𝑓(0) + 𝑓 ′ (0)𝑥 + + ⋯ ≈ 0 + 0 + 𝑥 2 (Maclaurin series)
= lim 2 2! 2
x→0 x (1 + cosx) 1−cos3 x
sin2 x 1 lim
x→0 x2
= lim 2 ∗ 3 2
x→0 x 1 + cosx 𝑥
2
1 = lim
= 12 ∗ x→0 x2
2 3
1 = lim
x→0 2
= 3
2 =
2



One-sided Limits
If f(x) approaches L1 as x approaches a from the Left, we write: 𝑙𝑖𝑚−𝑓(𝑥) = 𝐿1
𝑥→𝑎
If f(x) approaches L1 as x approaches a from the Left, we write: 𝑙𝑖𝑚+𝑓(𝑥) = 𝐿2
𝑥→𝑎

Relationship between one-sided and two-sided limits
If both 𝑙𝑖𝑚− 𝑓(𝑥) and 𝑙𝑖𝑚+ 𝑓(𝑥) exist and 𝑙𝑖𝑚−𝑓(𝑥) = 𝑙𝑖𝑚+ 𝑓(𝑥), then 𝑙𝑖𝑚𝑓(𝑥) exists and 𝑙𝑖𝑚𝑓(𝑥) = 𝑙𝑖𝑚− 𝑓(𝑥) = 𝑙𝑖𝑚+ 𝑓(𝑥).
𝑥→𝑎 𝑥→𝑎 𝑥→𝑎 𝑥→𝑎 𝑥→𝑎 𝑥→𝑎 𝑥→𝑎 𝑥→𝑎
The converse holds true as well.

Removal of Modulus Signs
|𝑥−3| |𝑥−3|
Find: 𝑙𝑖𝑚 Find: 𝑙𝑖𝑚
𝑥→3+ |𝑥|−|𝑥−6| 𝑥→3− |𝑥|−|𝑥−6|
Solutions. Solutions.
|𝑥−3| |𝑥−3|
𝑙𝑖𝑚 𝑙𝑖𝑚
𝑥→3+ |𝑥|−|𝑥−6| 𝑥→3− |𝑥|−|𝑥−6|
𝑥−3 −(𝑥−3)
= 𝑙𝑖𝑚 (removal of modulus sign) = 𝑙𝑖𝑚 (removal of modulus sign)
𝑥→3+ 𝑥−(−(𝑥−6)) 𝑥→3− 𝑥−(−(𝑥−6))
𝑥−3 −(𝑥−3)
= 𝑙𝑖𝑚 = 𝑙𝑖𝑚
𝑥→3+ 𝑥+𝑥−6 𝑥→3− 𝑥+𝑥−6
1 1
= 𝑙𝑖𝑚 = 𝑙𝑖𝑚 −
𝑥→3+ 2 𝑥→3− 2
1 1
= =−
2 2

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