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Midterm

MATH 1200 Midterm: Review for Midterm 2
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3 Pages
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Fall 2016

Department
Mathematics
Course Code
MATH 1200
Professor
Matthew Demers
Study Guide
Midterm

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Keefer Rourke
November 19, 2016
MATH 1200
Dr. Matt Demers
Calculus midterm 2
Formal definition of a limit
For any limit limxaf(x) = L, we can define:
ϵ > 0δ > 0, s.t. if 0<|xa|< δ, then |f(x)L|< ϵ
Limits to infinity
We can construct an argument similar to δϵ, for limits to infinity. The argument is as follows:
case 1:
lim
xaf(x) =
N > 0δ > 0, s.t. if 0<|xa|< δ, then f(x)> N.
where you start with f(x)> N, and work towards 0<|xa|< δ.
Note, if the limit is −∞, then use N < 0.
case 2:
lim
x→∞ f(x) = L
ϵ > 0M > 0s.t. if x > M, then |f(x)L|< ϵ
where you start with f|(x)L|< ϵ, and work towards M > 0.
Note, if approaching −∞, then use x < M and M < 0.
case 3:
lim
x→∞ f(x) =
N > 0M > 0, s.t. if x > M, then f(x)> N.
Note, if approaching −∞, then use x < M and M < 0— if the limit is −∞, then use N < 0.
Intermediate value theorem
If f(x)is continuous on [a, b], then for any y on that interval value, there must be x value that satisfies it.
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Description
Keefer Rourke MATH 1200 November 19, 2016 Dr. Matt Demers Calculus midterm 2 Formal definition of a limit For any limit ▯x▯a ▯(▯) = ▯, we can define: 8ϵ ▯ 09▯ ▯ 0▯▯▯▯▯ if 0 ▯ j▯ ▯ ▯j ▯ ▯▯ then j▯(▯) ▯ ▯j ▯ ϵ Limits to infinity We can construct an argument similar to ▯ ▯ ϵ, for limits to infinity. The argument is as follows: case 1: x▯a ▯(▯) = 1 8▯ ▯ 09▯ ▯ 0▯▯▯▯▯ if 0 ▯ j▯ ▯ ▯j ▯ ▯▯ then ▯(▯) ▯ ▯▯ where you start with ▯(▯) ▯ ▯, and work towards 0 ▯ j▯ ▯ ▯j ▯ ▯. Note, if the limit is ▯1, then use ▯ ▯ 0. case 2: lim ▯(▯) = ▯ x▯▯ 8ϵ ▯ 09▯ ▯ 0▯▯▯▯ if ▯ ▯ ▯▯ then j▯(▯) ▯ ▯j ▯ ϵ where you start with ▯j(▯) ▯ ▯j ▯ ϵ, and work towards ▯ ▯ 0. Note, if approaching ▯1, then use ▯ ▯ ▯ and ▯ ▯ 0. case 3: lim ▯(▯) = 1 x▯▯ 8▯ ▯ 09▯ ▯ 0▯▯▯▯▯ if ▯ ▯ ▯▯ then ▯(▯) ▯ ▯▯ Note, if approaching ▯1, then use ▯ ▯ ▯ and ▯ ▯ 0 — if the limit is ▯1, then use ▯ ▯ 0. Intermediate value theorem If ▯(▯) is continuous on [▯▯▯], then for any y on that interval value, there must be x value that satisfies it. 1 Keefer Rourke MATH 1200 November 19, 2016 Dr. Matt Demers Extreme value theorem If ▯(▯) is continuous on [▯▯▯], then 9 some ▯ and ▯ such that ▯(▯) ▯ ▯(▯) ▯ ▯(▯)▯8▯ 2 [▯▯▯]. I.e. there must be a maximum and minimum height somewhere on that interval. First principles ▯ ▯(▯ + ▯) ▯ ▯(▯) ▯▯ ▯(▯) = h▯0 ▯ Continuity and differentiability Differentiability implies continuity, but continuity does not imply differentiability. For instance, functions may have cusps or corners which do not have defined slopes at those p
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