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•Function
- is a binary relation over two sets that associates
every element of the first set, to exactly one
element of the second set.
•Evaluating a Function
-Means replacing the variable in the function.
Evaluate the following functions at x = 1.5.
Evaluate the following functions at x = 1.5.
Evaluate the following functions at x = 1.5.
Evaluate the following functions at x = 1.5.
Evaluate the following functions at x = 1.5.
Replace the value of x r(a+h) =
Note: Use FOIL(First,Outer,Inner,Last) method for
(a+h)(a+h) =
r(a+h) =
r(a+h) =
5
5 0
5 0 -3
5 0 -3 -4
5 0 -3 -4 -3
5 0 -3 -4 -3 0
5 0 -3 -4 -3 0 5
The analysis of the behaviour of a function as it
approaches some point (which may or may not be in
the domain of the function!).
This comes up in the real world all the time: any time
a model uses “ideal” conditions, we are looking at a
limit.
Is the value that a function (or sequence)
"approaches" as the input (or index) "approaches"
some value.
The fact that a function f approaches the
limit L as x approaches a is sometimes denoted by a
right arrow (→), as in:
When we evaluate a lim
𝑥→𝑎
𝑓(𝑥) we do one of the
following.
1. Find the limit value L in simplified form:
We write:
When we evaluate a lim
𝑥→𝑎
𝑓(𝑥) we do one of the
following.
2. When the limit is infinity (∞) or negative infinity
(−∞): We write:
When we evaluate a lim
𝑥→𝑎
𝑓(𝑥) we do one of the
following.
3. When the limit “Does Not Exist (DNE)” in some
other way,we write:
When we evaluate a lim
𝑥→𝑎
𝑓(𝑥) we do one of the
following.
If we say the limit is ∞ or −∞ , the limit is still not
exist. Think of ∞ or −∞ as “special cases of DNE”.
exists
The limit is a real number, L.
does not exist “DNE”
 -
x  a
Let f(x) = 3x2 – x + 1, evaluate 𝐥𝐢𝐦
𝒙→𝟏
𝒇(𝒙)
𝐥𝐢𝐦
𝒙→1
)𝒇(𝒙 = 𝒍𝒊𝒎
𝒙→𝟏
𝒇(𝟑𝐱 𝟐
− 𝐱 + 𝟏
= 𝟑(𝟏) 𝟐
−(𝟏) + 𝟏
= 𝟑(𝟏)−𝟏 +𝟏
= 𝟑 −𝟏 +𝟏
= 𝟑
𝐥𝐢𝐦
𝒙→1
)𝒇(𝒙 = 3
Let f(x) = 12x2 – 3x + 8, evaluate 𝐥𝐢𝐦
𝒙→𝟐
𝒇(𝒙)
𝐥𝐢𝐦
𝒙→2
)𝒇(𝒙 = 𝒍𝒊𝒎
𝒙→𝟐
𝒇(𝟏𝟐𝐱 𝟐
− 𝟑𝐱 + 𝟖
= 𝟏𝟐(𝟐) 𝟐
−𝟑(𝟐) + 𝟖
= 𝟏𝟐(𝟒) −𝟔 +𝟖
𝐥𝐢𝐦
𝒙→2
)𝒇(𝒙 = 50
= 𝟒𝟖 +𝟐
= 𝟓𝟎
Let f(x) =
𝟔+𝟒𝒕
𝒕 𝟐+𝟏
, evaluate 𝐥𝐢𝐦
𝒙→−𝟑
𝒇(𝒙)
𝐥𝐢𝐦
𝒙→−3
)𝒇(𝒙 = 𝒍𝒊𝒎
𝒙→−𝟑
𝟔 + 𝟒𝒕
𝒕 𝟐 + 𝟏
=
𝟔 + 𝟒(−𝟑)
(−𝟑) 𝟐+𝟏𝐥𝐢𝐦
𝒙→2
)𝒇(𝒙 =
−𝟑
𝟓
=
𝟔 − 𝟏𝟐
𝟗 + 𝟏
=
−𝟔
𝟏𝟎
=
−𝟑
𝟓
Basic Calculus Lesson 1
Basic Calculus Lesson 1

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Basic Calculus Lesson 1

  • 1.
  • 2.
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  • 4. •Function - is a binary relation over two sets that associates every element of the first set, to exactly one element of the second set. •Evaluating a Function -Means replacing the variable in the function.
  • 5. Evaluate the following functions at x = 1.5.
  • 6. Evaluate the following functions at x = 1.5.
  • 7. Evaluate the following functions at x = 1.5.
  • 8. Evaluate the following functions at x = 1.5.
  • 9. Evaluate the following functions at x = 1.5.
  • 10. Replace the value of x r(a+h) = Note: Use FOIL(First,Outer,Inner,Last) method for (a+h)(a+h) = r(a+h) = r(a+h) =
  • 11.
  • 12.
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  • 15.
  • 16.
  • 17. 5
  • 18. 5 0
  • 20. 5 0 -3 -4
  • 21. 5 0 -3 -4 -3
  • 22. 5 0 -3 -4 -3 0
  • 23. 5 0 -3 -4 -3 0 5
  • 24.
  • 25. The analysis of the behaviour of a function as it approaches some point (which may or may not be in the domain of the function!). This comes up in the real world all the time: any time a model uses “ideal” conditions, we are looking at a limit.
  • 26. Is the value that a function (or sequence) "approaches" as the input (or index) "approaches" some value.
  • 27. The fact that a function f approaches the limit L as x approaches a is sometimes denoted by a right arrow (→), as in:
  • 28. When we evaluate a lim 𝑥→𝑎 𝑓(𝑥) we do one of the following. 1. Find the limit value L in simplified form: We write:
  • 29. When we evaluate a lim 𝑥→𝑎 𝑓(𝑥) we do one of the following. 2. When the limit is infinity (∞) or negative infinity (−∞): We write:
  • 30. When we evaluate a lim 𝑥→𝑎 𝑓(𝑥) we do one of the following. 3. When the limit “Does Not Exist (DNE)” in some other way,we write:
  • 31. When we evaluate a lim 𝑥→𝑎 𝑓(𝑥) we do one of the following. If we say the limit is ∞ or −∞ , the limit is still not exist. Think of ∞ or −∞ as “special cases of DNE”. exists The limit is a real number, L. does not exist “DNE”  - x  a
  • 32. Let f(x) = 3x2 – x + 1, evaluate 𝐥𝐢𝐦 𝒙→𝟏 𝒇(𝒙) 𝐥𝐢𝐦 𝒙→1 )𝒇(𝒙 = 𝒍𝒊𝒎 𝒙→𝟏 𝒇(𝟑𝐱 𝟐 − 𝐱 + 𝟏 = 𝟑(𝟏) 𝟐 −(𝟏) + 𝟏 = 𝟑(𝟏)−𝟏 +𝟏 = 𝟑 −𝟏 +𝟏 = 𝟑 𝐥𝐢𝐦 𝒙→1 )𝒇(𝒙 = 3
  • 33. Let f(x) = 12x2 – 3x + 8, evaluate 𝐥𝐢𝐦 𝒙→𝟐 𝒇(𝒙) 𝐥𝐢𝐦 𝒙→2 )𝒇(𝒙 = 𝒍𝒊𝒎 𝒙→𝟐 𝒇(𝟏𝟐𝐱 𝟐 − 𝟑𝐱 + 𝟖 = 𝟏𝟐(𝟐) 𝟐 −𝟑(𝟐) + 𝟖 = 𝟏𝟐(𝟒) −𝟔 +𝟖 𝐥𝐢𝐦 𝒙→2 )𝒇(𝒙 = 50 = 𝟒𝟖 +𝟐 = 𝟓𝟎
  • 34. Let f(x) = 𝟔+𝟒𝒕 𝒕 𝟐+𝟏 , evaluate 𝐥𝐢𝐦 𝒙→−𝟑 𝒇(𝒙) 𝐥𝐢𝐦 𝒙→−3 )𝒇(𝒙 = 𝒍𝒊𝒎 𝒙→−𝟑 𝟔 + 𝟒𝒕 𝒕 𝟐 + 𝟏 = 𝟔 + 𝟒(−𝟑) (−𝟑) 𝟐+𝟏𝐥𝐢𝐦 𝒙→2 )𝒇(𝒙 = −𝟑 𝟓 = 𝟔 − 𝟏𝟐 𝟗 + 𝟏 = −𝟔 𝟏𝟎 = −𝟑 𝟓