# Feg 1034 Calculus & Analysis 1

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Autor:   •  December 6, 2017  •  Research Paper  •  1,195 Words (5 Pages)  •  148 Views

Page 1 of 5

FEG 1034        Calculus 1        Chapter 02[pic 1]

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Chapter 2

Function and limit

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FEG 1034 Calculus & Analysis 1

FEG 1034        Calculus 1        Chapter 05[pic 4]

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What thing will come out in Final

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2.1 Limit

• Vertical Asymptotes

• Horizontal Asymptotes

2.2 Inverse Function

• Characteristic of inverse function

• Horizontal test
• Second derivative test

2.3 Derivative of inverse function

FEG 1034 Calculus & Analysis 1        2

FEG 1034        Calculus 1        Chapter 02[pic 7]

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2.1 Limit

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= 3−1 and        =  2 +        + 1

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−1

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Both graph are the same, but just that can’t include = 1, which is not continuous when = 1.

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FEG 1034 Calculus & Analysis 1        3

FEG 1034        Calculus 1        Chapter 02[pic 15]

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2.1.1 Vertical Asymptotes

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How to determine the vertical asymptotes?

• Check the existence of denominator (must be simplify)

• Check the value to make denominator become zero, hence make the function become undefined.

For this example, the vertical asymptotes is in        = 2.

FEG 1034 Calculus & Analysis 1        4

FEG 1034        Calculus 1        Chapter 02[pic 18]

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Example 1 – find vertical asymptotes

Find the vertical asymptotes for the functions.

a).        = 1

b).        = 1

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FEG 1034 Calculus & Analysis 1        5

FEG 1034        Calculus 1        Chapter 02[pic 21]

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2.1.2 Horizontal Asymptotes

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How to determine the horizontal asymptotes?

• If the limit at infinite is defined, then there is the horizontal asymptotes.

=

For this example,

 3 lim = −0 3 = 0 →−∞  −2 −∞

lim        3        = 0

[pic 24]

→+∞  −2

3

[pic 25][pic 26][pic 27][pic 28][pic 29]

+∞ = 0

horizontal asymptotes

at        =  .

FEG 1034 Calculus & Analysis 1        6

FEG 1034        Calculus 1        Chapter 02[pic 30]

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Example 2 – find horizontal asymptotes

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Find the horizontal asymptotes for the functions.

 2 + 1 2 1 1 = = 1 + = 1 − + 2 − 1 2 − 1 + 1 − 1 = lim 2 + 1 = 1 + lim − 1 + 1 = 1 − 1 + 1 − 1 →−∞  2 →−∞ lim 2 + 1 = 1 + lim − 1 + 1 = 1 − 1 + 1 − 1 →+∞  2 →+∞

[pic 33][pic 34]

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