Math119_Road To success_Booklet_Free_Trial

[MATH 119 BOOKLET]

INTRODUCTION

This booklet is for those who are taking Math119 (Calculus with Analytic Geometry).

What does the booklet offer to their students?

1.

2.

3.

4.

5.

6.

Sufficient way to read the calculus subjects.

Organized way to understand calculus theorems and some proves as well as some necessary rules.

Sample examples to understand how to solve the related problems Some suggested problems to practice Appendix which contains functions and their graphs Geometric formulas Short answer key of the suggested problems at the end of the booklet as well as full solution in the answer key booklet.

đ?‘…đ?‘’đ?‘Žđ?‘‘ ⇢ đ?‘ˆđ?‘›đ?‘‘đ?‘’đ?‘&#x;đ?‘ đ?‘Ąđ?‘Žđ?‘›đ?‘‘ ⇢ đ?‘ƒđ?‘&#x;đ?‘Žđ?‘?đ?‘Ąđ?‘–đ?‘?đ?‘’ ⇢ đ??´đ??´ What are the external booklets offered?

1.

2.

3.

4.

Notebook Booklet (This booklet) Answer key Booklet Sample midterms and Final exam Booklet Brief Booklet

Note: Math 119 booklets can be found online via â„Žđ?‘Ąđ?‘Ąđ?‘?://đ?‘šđ?‘’đ?‘Ąđ?‘˘. đ?‘Žđ?‘?đ?‘Žđ?‘‘đ?‘’đ?‘šđ?‘–đ?‘Ž. đ?‘’đ?‘‘đ?‘˘/đ??ťđ?‘Žđ?‘ đ?‘ đ?‘Žđ?‘›đ?‘–đ?‘›đ?‘…đ?‘–đ?‘ â„Žđ?‘Ž To have chance of having mock/sample exams as well as other online extra recourses and support, you may join â„Žđ?‘Ąđ?‘Ąđ?‘?đ?‘ ://đ?‘šđ?‘Žđ?‘Ąâ„Ž. đ?‘”đ?‘›đ?‘œđ?‘šđ?‘–đ?‘œ. đ?‘?đ?‘œđ?‘š

If you have any inquiries, please do not hesitate to contact me via â„Žđ?‘Žđ?‘ đ?‘ đ?‘Žđ?‘›đ?‘–đ?‘›. đ?‘&#x;đ?‘–đ?‘ â„Žđ?‘Ž @ đ?‘”đ?‘šđ?‘Žđ?‘–đ?‘™. đ?‘?đ?‘œđ?‘š I hope such a work helps you to reach AA letter grade.

Best Regards,

Copyright Š 2019 Road To success All Rights Reserved. Email: hassanin.risha@gmail.com Website: https://math.gnomio.com

1

Contents Section 1: How to solve limit problems................................................................................................................................. 4 Section 2: Continuity and Differentiation ........................................................................................................................... 16 Section 3: Derivatives........................................................................................................................................................... 21 Section 4: Tangent Lines and Their Slope ............................................................................................................................ 29 Section 5: Prove Rules with their conditions ...................................................................................................................... 33 Section 6: Related Rates ...................................................................................................................................................... 38 Section 7: Indeterminate Forms .......................................................................................................................................... 40 Section 8: Curve Sketching Steps ......................................................................................................................................... 45 Section 9: Extreme-Value Problem ...................................................................................................................................... 57 Section 10: Linear Approximations ..................................................................................................................................... 61 Section 11: Sums and Sigma Notation................................................................................................................................. 62 Section 12: The Definite Integral ........................................................................................................................................ 65 Section 13: Properties of the Definite Integral ................................................................................................................... 67 Section 14: The Fundamental Theorem of Calculus [F.T.C] ................................................................................................ 67 Section 15.1: Integral Properties ......................................................................................................................................... 69 Section 15.2: Substitution .................................................................................................................................................... 69 Section 15.3: Integration by Parts ....................................................................................................................................... 72 Section 15.4: Partial Fractions ............................................................................................................................................. 75 Section 15.5: Trig Substitutions ........................................................................................................................................... 78 Section 16: Improper Integrals ............................................................................................................................................ 80 Section 17: Areas of Plane Regions ..................................................................................................................................... 87 Section 18: Volumes by Slicing-Solids of Revolution .......................................................................................................... 89 Section 19: Arc Length ......................................................................................................................................................... 93 Section 20: Polar Coordinates and Polar Curves ................................................................................................................. 93 Answer key: .......................................................................................................................................................................... 95

Copyright Š 2019 Road To success All Rights Reserved. Email: hassanin.risha@gmail.com Website: https://math.gnomio.com

2

Booklet Sections (Road to Success) Section 1: How to solve limit problems Section 2: Continuity and Differentiation Section 3: Derivatives

Section 4: Tangent Lines and Their Slope Section 5: Prove Rules with their conditions

Book Cheaters (A Complete Course Calculus) 1.2 Limits of Functions 1.3 Limits at Infinity and Infinite Limits 1.4 Continuity Ch 2: Differentiation 3.1 Inverse Functions 3.2 Exponential and Logarithmic Functions 3.3 The Natural Logarithm and Exponential 3.5 The Inverse Trigonometric Functions 3.6 Hyperbolic Functions 2.1 Tangent Lines and Their Slope

Section 6: Related Rates

1.5 The Formal Definition of Limit 2.8 The Mean-Value Theorem 1.4 Continuity (Intermediate value theorem) 4.1 Related Rates

Section 7: Indeterminate Forms

4.3 Indeterminate Forms

Section 8: Curve Sketching Steps

4.4 Extreme Values 4.5 Concavity and Inflections

Section 9: Extreme-Value Problem

4.6 Sketching the Graph of a Function 4.8 Extreme-Value Problems

Section 10: Linear Approximations

4.9 Linear Approximations

Section 11: Sums and Sigma Notation Section 12: The Definite Integral

5.1 Sums and Sigma Notation 5.2 Areas as Limits of Sums 5.3 The Definite Integral

Section 13: Properties of the Definite Integral

5.4 Properties of the Definite Integral

Section 14: The Fundamental Theorem of Calculus

5.5 The Fundamental Theorem of Calculus

Section 15.1: Integral Properties

5.3 The Definite Integral

Section 15.2: Substitution

5.6 The Method of Substitution

Section 15.3: Integration by Parts

6.1 Integration by Parts

Section 15.4: Partial Fractions

6.2 Integrals of Rational Functions

Section 15.5: Trig Substitutions

6.3 Inverse Substitutions

Section 16: Improper Integrals

6.5 Improper Integrals

Section 17: Areas of Plane Regions

5.7 Areas of Plane Regions

Section 18: Volumes by Slicing-Solids of Revolution

7.1 Volumes by Slicing-Solids of Revolution

Section 19: Arc Length

7.3 Arc Length

Section 20: Polar Coordinates and Polar Curves

8.5 Polar Coordinates and Polar Curves

Copyright Š 2019 Road To success All Rights Reserved. Email: hassanin.risha@gmail.com Website: https://math.gnomio.com

3

Section 1: How to solve limit problems Strategy to Calculate Limits

Hint 1: (Very Important!) l’Hˆopital’s theorem (section 7) should not be used at this part since it involves taking derivatives. In the exam, if you are NOT asked to use this theorem, then Instructors will NOT give any credit for any answer that includes such a method.

To compute lim đ?‘“(đ?‘Ľ) = đ??ż đ?‘œđ?‘&#x; lim đ?‘”(đ?‘Ľ) = đ?‘€ đ?‘Ľâ†’đ?’‚

I.

�→�

You need to know the limit rules: •

Limit of a sum or difference: lim [đ?‘“(đ?‘Ľ) Âą đ?‘”(đ?‘Ľ)] = đ??ż Âą đ?‘€

•

Limit of a product: lim [đ?‘“(đ?‘Ľ)đ?‘”(đ?‘Ľ)] = đ??żđ?‘€

•

Limit of a quotient: lim

•

Limit of a power: lim [ đ?‘“(đ?‘Ľ)]đ?‘› = đ??żđ?‘› , đ?‘› ∈ ℤ (đ?‘–đ?‘›đ?‘Ąđ?‘’đ?‘”đ?‘’đ?‘&#x;)

II.

�→�

�→�

đ?‘“(đ?‘Ľ)

� → � �(�)

=

đ??ż

đ?‘¤â„Žđ?‘’đ?‘&#x;đ?‘’ đ?‘€ ≠0

đ?‘€

Hint 2: −1 ≤ sin(đ?‘Ľ) ≤ 1 −1 ≤ cos (đ?‘Ľ) ≤ 1 0 ≤ |sin(đ?‘Ľ)| ≤ 1 0 ≤ |cos (đ?‘Ľ)| ≤ 1 0 ≤ sin2 (đ?‘Ľ) ≤ 1 0 ≤ cos 2 (đ?‘Ľ) ≤ 1

�→�

How to solve the limit:

Step 1: Try to plug the value of � directly into the function. •

If we get a number (đ??ż đ?‘Žđ?‘›đ?‘‘ đ?‘€ ∈ â„?), then we are done!

đ?‘Ľ 2 − 2đ?‘Ľ − 3 −4 đ??¸đ?‘Ľđ?‘Žđ?‘šđ?‘?đ?‘™đ?‘’: lim = = −2 đ?‘Ľâ†’1 đ?‘Ľ+1 2 •

However, the limit/value is undeďŹ ned, having the form 0

∞

0

∞

đ?‘Ąâ„Žđ?‘’đ?‘›

( [ ] , [ ] , đ?‘œđ?‘&#x; [0 ∗ ∞]) (đ?’“đ?’†đ?’‚đ?’… đ?’‰đ?’Šđ?’?đ?’• đ?&#x;?) →

Follow step 2

Step 2: If the limit/value is undeďŹ ned, then we need to simplify the expression. SimpliďŹ cation can involve any number of techniques including but certainly not limited to the 7 given techniques:

1.

Hint 3: sin đ?‘Ľ cos đ?‘Ľ lim = lim =0 đ?‘Ľâ†’∞ đ?‘Ľ đ?‘Ľâ†’∞ đ?‘Ľ sin đ?‘Ľ lim =1 đ?‘Ľâ†’0 đ?‘Ľ Hint 4: 1 1 lim = 0 lim+ = +∞ đ?‘Ľâ†’∞đ?‘Ľ đ?‘Ľâ†’0 đ?‘Ľ 1 1 lim− = −∞ lim = đ??ˇ. đ?‘ . đ??¸ đ?‘Ľâ†’0 đ?‘Ľâ†’0 đ?‘Ľ đ?‘Ľ Hint 5: đ?‘’ ≈ 2.7 ln 1 = 0 ln đ?‘’ = 1 ln 0 = −∞ ln ∞ = ∞ đ?‘’ ∞ = ∞ đ?‘’ −∞ = 0 đ??śâ„Žđ?‘’đ?‘?đ?‘˜ đ?‘ đ?‘œđ?‘Ąđ?‘’ đ?‘‰đ??ź

Factoring (chapter 1.2)

lim

đ?‘Ľ 2 +2đ?‘Ľâˆ’3

=

(đ?‘Ľ+3)(đ?‘Ľâˆ’1) đ?‘Ľâˆ’1

=đ?‘Ľ+3=4

đ?‘“đ?‘Žđ?‘?đ?‘Ąđ?‘œđ?‘&#x;đ?‘–đ?‘›đ?‘” đ?‘Ąâ„Žđ?‘’ đ?‘™đ?‘–đ?‘šđ?‘–đ?‘Ą đ?‘Ąđ?‘œ đ?‘”đ?‘’đ?‘Ą đ?‘&#x;đ?‘–đ?‘‘ đ?‘œđ?‘“ (đ?‘Ľ − 1)

�→1

đ?‘Ľâˆ’1

2.

Use of the absolute functions in limits (chapter 1.2)

Theorem: A function đ?‘“(đ?‘Ľ) has limit đ??ż at đ?‘Ľ = đ?‘Ž if and only if it has both the right and left limits there and these one-sided limits are both equal to đ??ż: lim đ?‘“(đ?‘Ľ) = đ??ż â&#x;ş

�→�

lim đ?‘“(đ?‘Ľ) = lim+ đ?‘“(đ?‘Ľ) = đ??ż

đ?‘Ľ → đ?’‚−

�→�

đ?‘Ľ2 − 4 (đ?‘Ľ − 2)(đ?‘Ľ + 2) (đ?‘Ľ − 2) −4 = lim + = lim + = =2 2 đ?‘Ľ −4 đ?‘Ľ → −2 đ?‘Ľ(đ?‘Ľ + 2) đ?‘Ľ → −đ?&#x;? đ?‘Ľ → −2 đ?‘Ľ(đ?‘Ľ + 2) đ?‘Ľ −2 đ?‘’đ?‘Ľ: lim = đ?‘Ľ → −2 đ?‘Ľ |đ?‘Ľ + 2| (đ?‘Ľ − 2) −4 đ?‘Ľ2 − 4 (đ?‘Ľ − 2)(đ?‘Ľ + 2) lim − = lim − = lim − = = −2 đ?‘Ľ → −2 −đ?‘Ľ(đ?‘Ľ + 2) −đ?‘Ľ 2 {đ?‘Ľ → −2 −đ?‘Ľ (đ?‘Ľ + 2) đ?‘Ľ → −đ?&#x;? lim +

đ?‘“đ?‘&#x;đ?‘œđ?‘š (1) đ?‘Žđ?‘›đ?‘‘ (2), đ?‘Šđ?‘’ đ?‘?đ?‘Žđ?‘› đ?‘?đ?‘œđ?‘›đ?‘?đ?‘˘đ?‘™đ?‘‘đ?‘’ đ?‘Ąâ„Žđ?‘Žđ?‘Ą lim

đ?‘Ľ → −2

→ (1) → (2)

đ?‘Ľ2 − 4 đ??ˇđ?‘œđ?‘’đ?‘ đ?‘ đ?‘œđ?‘Ą đ??¸đ?‘Ľđ?‘–đ?‘ đ?‘Ą đ?‘†đ?‘–đ?‘›đ?‘?đ?‘’ đ?‘?đ?‘œđ?‘Ąâ„Ž đ?‘&#x;đ?‘–đ?‘”â„Žđ?‘Ą đ?‘Žđ?‘›đ?‘‘ đ?‘™đ?‘’đ?‘“đ?‘Ą đ?‘ đ?‘–đ?‘‘đ?‘’ đ?‘Žđ?‘&#x;đ?‘’ đ?‘‘đ?‘–đ?‘“đ?‘“đ?‘’đ?‘&#x;đ?‘’đ?‘›đ?‘Ą. đ?‘Ľ |đ?‘Ľ + 2|

Check general note III (3rd one) to learn more about the absolute functions in limit

Copyright Š 2019 Road To success All Rights Reserved. Email: hassanin.risha@gmail.com Website: https://math.gnomio.com

4

3.

Using Squeeze Theorem (chapter 1.2)

Theorem: suppose that đ?‘”(đ?‘Ľ) ≤ đ?‘“(đ?‘Ľ) ≤ â„Ž(đ?‘Ľ) holds for all đ?‘Ľ in some open interval containing a, except possibly at đ?‘Ľ = đ?‘Ž itself. Suppose also that lim đ?‘”(đ?‘Ľ) = lim â„Ž(đ?‘Ľ) = đ??ż đ?‘Ľâ†’đ?‘Ž

�→�

Then lim đ?‘“(đ?‘Ľ) = đ??ż đ?‘Žđ?‘ đ?‘™đ?‘œ. đ?‘†đ?‘–đ?‘šđ?‘–đ?‘™đ?‘Žđ?‘&#x; đ?‘ đ?‘Ąđ?‘Žđ?‘Ąđ?‘’đ?‘šđ?‘’đ?‘›đ?‘Ąđ?‘ â„Žđ?‘œđ?‘™đ?‘‘ đ?‘“đ?‘œđ?‘&#x; đ?‘™đ?‘’đ?‘“đ?‘Ą đ?‘Žđ?‘›đ?‘‘ đ?‘&#x;đ?‘–đ?‘”â„Žđ?‘Ą đ?‘™đ?‘–đ?‘šđ?‘–đ?‘Ąđ?‘ . (read hint 2) đ?‘Ľâ†’đ?‘Ž

2sin(đ?‘Ľ 3 )

đ?‘’đ?‘Ľ: lim

đ?‘Ľ 3 +1

đ?‘Ľâ†’∞

lim

−2

đ?‘Ľâ†’∞

đ?‘Ľ 3 +1

2sin(đ?‘Ľ 3 ) =0 đ?‘Ľâ†’∞ đ?‘Ľ 3 + 1

2 −2 = lim 3 =0 3 đ?‘Ľâ†’∞ đ?‘Ľ + 1 đ?‘Ľâ†’∞ đ?‘Ľ + 1

đ?‘ đ?‘–đ?‘›đ?‘?đ?‘’ 4.

đ?‘ đ?‘–đ?‘›đ?‘?đ?‘’ − 1 ≤ sin(đ?‘Ľ 3 ) ≤ 1 →

→

lim

∴ lim

≤ lim

2sin(đ?‘Ľ 3 )

đ?‘Ľâ†’∞

2

≤ lim

đ?‘Ľ 3 +1

đ?‘Ľâ†’∞ đ?‘Ľ 3 +1

đ??ľđ?‘Ś đ?‘†đ?‘žđ?‘˘đ?‘’đ?‘’đ?‘§đ?‘’ đ?‘‡â„Žđ?‘’đ?‘œđ?‘&#x;đ?‘’đ?‘š

Use of the limit rules Finding a common denominator (chapter 1.3) 2đ?‘Ľ 4

lim

2đ?‘Ľ + √81đ?‘Ľ 6 + đ?‘Ľ 5 − 3đ?‘Ľ + 7 5đ?‘Ľ + 8 + √đ?‘Ľ 3 + đ?‘Ľ − 2

đ?‘Ľâ†’∞

= lim

3 đ?‘Ľ2

đ?‘Ľâ†’∞

4 81đ?‘Ľ 6 đ?‘Ľ 5 3đ?‘Ľ 7 +√ 6 + 6− 6+ 6 đ?‘Ľ đ?‘Ľ đ?‘Ľ đ?‘Ľ

5đ?‘Ľ 3

√ 3+

đ?‘Ľ2 1

đ?‘Ľ3 đ?‘Ľ 2 + 3− 3 3 đ?‘Ľ đ?‘Ľ đ?‘Ľ

8

+

đ?‘Ľ2

4

(2đ?‘Ľ −2 + √81 + đ?‘Ľ −1 − 3đ?‘Ľ −5 + 7đ?‘Ľ −6 ) lim

1 đ?‘Ľâ†’∞ (5đ?‘Ľ −2

+

(đ?‘¤â„Žđ?‘’đ?‘&#x;đ?‘’ lim

1

đ?‘Ľâ†’∞đ?‘Ľ

=

3 8đ?‘Ľ −2

4

4

(0 + √81 + 0 − 3 ∗ 0 + 7 ∗ 0) (0 + 8 ∗ 0 + √1 + 0 − 2 ∗ 0)

+ √1 + đ?‘Ľ −2 − 2đ?‘Ľ −3 )

=

√81 =3 1

= 0) 3

•

Check the highest power in the denominator then factor it out from both denominator and nominator (đ?‘Ľ 2 )

•

The general formula of lim

1

đ?‘Ľâ†’∞ đ?‘Ľ

đ?‘–đ?‘ lim

1

đ?‘Ľ → ∞ đ?‘Ľđ?‘?

=0

đ?&#x2018;¤â&#x201E;&#x17D;đ?&#x2018;&#x2019;đ?&#x2018;&#x;đ?&#x2018;&#x2019; 0 < đ?&#x2018;? < â&#x2C6;&#x17E;

â&#x20AC;Śâ&#x20AC;Ś

5.

Multiplying by the conjugate (chapter 1.3)

lim đ?&#x2018;Ľ â&#x2C6;&#x2019;

đ?&#x2018;Ľâ&#x2020;&#x2019;â&#x2C6;&#x17E;

â&#x2C6;&#x161;đ?&#x2018;Ľ 2

+đ?&#x2018;Ľ+2â&#x2C6;&#x2014;

â&#x2C6;&#x2019;đ?&#x2018;Ľâ&#x2C6;&#x2019;2

= lim

đ?&#x2018;Ľâ&#x2020;&#x2019;â&#x2C6;&#x17E; đ?&#x2018;Ľ+â&#x2C6;&#x161;đ?&#x2018;Ľ 2 +đ?&#x2018;Ľ+2

6. lim

đ?&#x2018;Ľ + â&#x2C6;&#x161;đ?&#x2018;Ľ 2 + đ?&#x2018;Ľ + 2

= lim

đ?&#x2018;Ľ + â&#x2C6;&#x161;đ?&#x2018;Ľ 2 + đ?&#x2018;Ľ + 2

đ?&#x2018;Ľ + â&#x2C6;&#x161;đ?&#x2018;Ľ 2 + đ?&#x2018;Ľ + 2

đ?&#x2018;Ľâ&#x2020;&#x2019;â&#x2C6;&#x17E;

đ?&#x2018;Ľ(â&#x2C6;&#x2019;1â&#x2C6;&#x2019;2đ?&#x2018;Ľ â&#x2C6;&#x2019;1 )

= lim

đ?&#x2018;Ľ 2 â&#x2C6;&#x2019; (đ?&#x2018;Ľ 2 + đ?&#x2018;Ľ + 2)

đ?&#x2018;Ľâ&#x2020;&#x2019;â&#x2C6;&#x17E; đ?&#x2018;Ľ(1+â&#x2C6;&#x161;1+đ?&#x2018;Ľ â&#x2C6;&#x2019;1 +2đ?&#x2018;Ľ â&#x2C6;&#x2019;2 )

=

â&#x2C6;&#x2019;1

đ?&#x2018;&#x2020;đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;?đ?&#x2018;&#x2019; lim

2

1

=0

đ?&#x2018;Ľâ&#x2020;&#x2019;â&#x2C6;&#x17E;đ?&#x2018;Ľ

Applying some memorized limit (chapter 2.5) (read hint 3) tan 5đ?&#x2018;Ľ

đ?&#x2018;Ľâ&#x2020;&#x2019;0 sin 3đ?&#x2018;Ľ

7.

= lim

sin 5đ?&#x2018;Ľ

đ?&#x2018;Ľâ&#x2020;&#x2019;0 cos 5đ?&#x2018;Ľ

â&#x2C6;&#x2014; lim

1

đ?&#x2018;Ľâ&#x2020;&#x2019;0 sin 3đ?&#x2018;Ľ

= lim

đ?&#x2018;Ľâ&#x2020;&#x2019;0

sin 5đ?&#x2018;Ľ 5đ?&#x2018;Ľ

â&#x2C6;&#x2014; lim

1

đ?&#x2018;Ľâ&#x2020;&#x2019;0 cos 5đ?&#x2018;Ľ

â&#x2C6;&#x2014; lim

3đ?&#x2018;Ľ

đ?&#x2018;Ľâ&#x2020;&#x2019;0 sin 3đ?&#x2018;Ľ

5

5

5

3

3

3

â&#x2C6;&#x2014; =1â&#x2C6;&#x2014;1â&#x2C6;&#x2014; 1 â&#x2C6;&#x2014; =

â&#x;š đ?&#x2018;&#x2020;đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;?đ?&#x2018;&#x2019; lim

sin đ?&#x2018;Ľ

đ?&#x2018;Ľâ&#x2020;&#x2019;0

đ?&#x2018;Ľ

=1

Using the deďŹ nition of the limit (chapter 2.7) đ?&#x2018;&#x201C;(đ?&#x2018;Ľ) â&#x2C6;&#x2019; đ?&#x2018;&#x201C;(đ?&#x2018;&#x17D;) = đ?&#x2018;&#x201C;â&#x20AC;˛(đ?&#x2018;&#x17D;) đ?&#x2018;Ľâ&#x2020;&#x2019;đ?&#x2018;&#x17D; đ?&#x2018;Ľâ&#x2C6;&#x2019;đ?&#x2018;&#x17D; lim

cos(đ?&#x2018;Ľ) â&#x2C6;&#x2019; 1 đ?&#x2018;&#x201C;(đ?&#x2018;Ľ) â&#x2C6;&#x2019; đ?&#x2018;&#x201C;(0) = lim = đ?&#x2018;&#x201C;â&#x20AC;˛(0) đ?&#x2018;Ľâ&#x2020;&#x2019;0 đ?&#x2018;Ľâ&#x2020;&#x2019; 0 đ?&#x2018;Ľ đ?&#x2018;Ľâ&#x2C6;&#x2019;0 lim

đ?&#x2018;¤â&#x201E;&#x17D;đ?&#x2018;&#x2019;đ?&#x2018;&#x;đ?&#x2018;&#x2019; đ?&#x2018;&#x201C;(đ?&#x2018;Ľ) = cos(đ?&#x2018;Ľ)

â&#x2C6;´ đ?&#x2018;&#x201C;â&#x20AC;˛(đ?&#x2018;Ľ) = â&#x2C6;&#x2019;đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A; đ?&#x2018;Ľ

â&#x2C6;´ đ?&#x2018;&#x201C;â&#x20AC;˛(0) = 0

General Notes about limits: I.â&#x2C6;&#x161;đ?&#x2018;Ľ 2 = |đ?&#x2018;Ľ| â&#x2020;&#x2019; { II.

đ?&#x2018;Ľ â&#x2C6;&#x2019;đ?&#x2018;Ľ

đ?&#x2018;Ľâ&#x2020;&#x2019;â&#x2C6;&#x17E; đ?&#x2018;Ľ â&#x2020;&#x2019; â&#x2C6;&#x2019;â&#x2C6;&#x17E;

Squeeze theorem is applicable for lim

đ?&#x2018;Ľâ&#x2020;&#x2019;â&#x2C6;&#x17E;

sin đ?&#x2018;Ľ đ?&#x2018;Ľ

1

= lim

đ?&#x2018;Ľâ&#x2020;&#x2019;0

sin(đ?&#x2018;Ľ) 1 đ?&#x2018;Ľ

= 0,

Not for lim

đ?&#x2018;Ľâ&#x2020;&#x2019;0

sin đ?&#x2018;Ľ đ?&#x2018;Ľ

1

= lim

đ?&#x2018;Ľâ&#x2020;&#x2019;â&#x2C6;&#x17E;

sin(đ?&#x2018;Ľ) 1 đ?&#x2018;Ľ

=1

Copyright Š 2019 Road To success All Rights Reserved. Email: hassanin.risha@gmail.com Website: https://math.gnomio.com

5

III.

For absolute questions, you need to know when you are going to use positive or negative sign. (example) đ?&#x2018;Ľâ&#x2020;&#x2019;2

Hints

|đ?&#x2018;Ľ â&#x2C6;&#x2019; 2|

{

|đ?&#x2018;Ľ â&#x2C6;&#x2019; 4|

{

đ?&#x2018;Ľâ&#x2C6;&#x2019;2 â&#x2C6;&#x2019;(đ?&#x2018;Ľ â&#x2C6;&#x2019; 2)

đ?&#x2018;Ľâ&#x2C6;&#x2019;4 â&#x2C6;&#x2019;(đ?&#x2018;Ľ â&#x2C6;&#x2019; 4)

đ?&#x2018;Ľ+2 { â&#x2C6;&#x2019;(đ?&#x2018;Ľ + 2)

|đ?&#x2018;Ľ + 2|

đ?&#x2018;Ľ>2 đ?&#x2018;Ľ<2

đ?&#x2018;Ľ â&#x2020;&#x2019; 2+ đ?&#x2018;Ľ>2

đ?&#x2018;Ľ â&#x2020;&#x2019; 2â&#x2C6;&#x2019; đ?&#x2018;Ľ<2

đ?&#x2018;Ľâ&#x2C6;&#x2019;2

â&#x2C6;&#x2019;(đ?&#x2018;Ľ â&#x2C6;&#x2019; 2)

â&#x2C6;&#x2019;(đ?&#x2018;Ľ â&#x2C6;&#x2019; 4)

â&#x2C6;&#x2019;(đ?&#x2018;Ľ â&#x2C6;&#x2019; 4)

đ?&#x2018;Ľ+2

đ?&#x2018;Ľ+2

â&#x;šđ?&#x2018;Ľ=2

đ?&#x2018;Ľ>4 đ?&#x2018;Ľ<4

â&#x;šđ?&#x2018;Ľ=4

đ?&#x2018;Ľ > â&#x2C6;&#x2019;2 đ?&#x2018;Ľ < â&#x2C6;&#x2019;2

â&#x;š đ?&#x2018;Ľ = â&#x2C6;&#x2019;2

IV. Look how to take the factor out from the equation: (this is just one way of factorizing) 3

1

3

5đ?&#x2018;Ľ + 8 â&#x2C6;&#x2019; â&#x2C6;&#x161;đ?&#x2018;Ľ 3 + đ?&#x2018;Ľ â&#x2C6;&#x2019; 2 = đ?&#x2018;Ľ 2 (5đ?&#x2018;Ľ â&#x2C6;&#x2019;2 + 8đ?&#x2018;Ľ â&#x2C6;&#x2019;2 â&#x2C6;&#x2019; â&#x2C6;&#x161;1 + đ?&#x2018;Ľ â&#x2C6;&#x2019;2 â&#x2C6;&#x2019; 2đ?&#x2018;Ľ â&#x2C6;&#x2019;3 ) Before factoring

5đ?&#x2018;Ľ

8

â&#x2C6;&#x161;đ?&#x2018;Ľ 3 + â&#x2039;Ż â&#x2C6;&#x2019; â&#x2039;Ż

â&#x2C6;&#x161;â&#x20AC;Ś + đ?&#x2018;Ľ â&#x2C6;&#x2019; â&#x2039;Ż

â&#x2C6;&#x161;â&#x20AC;Ś + â&#x2039;Ż â&#x2C6;&#x2019; 2

After factoring

1 5đ?&#x2018;Ľ â&#x2C6;&#x2019;2

3 8đ?&#x2018;Ľ â&#x2C6;&#x2019;2

â&#x2C6;&#x161;1 + â&#x2039;Ż â&#x2C6;&#x2019; â&#x2039;Ż

â&#x2C6;&#x161;â&#x20AC;Ś + đ?&#x2018;Ľ â&#x2C6;&#x2019;2 â&#x2C6;&#x2019; â&#x2039;Ż

â&#x2C6;&#x161;â&#x20AC;Ś + â&#x2039;Ż â&#x2C6;&#x2019; 2đ?&#x2018;Ľ â&#x2C6;&#x2019;3

3â&#x2C6;&#x2019;3=0

1 â&#x2C6;&#x2019; 3 = â&#x2C6;&#x2019;2

0 â&#x2C6;&#x2019; 3 = â&#x2C6;&#x2019;3

Power of đ?&#x2018;Ľ after 1â&#x2C6;&#x2019;

factoring

V. lim

đ?&#x2018;&#x201C;(đ?&#x2018;Ľ)

đ?&#x2018;Ľâ&#x2020;&#x2019; â&#x2C6;&#x17E; đ?&#x2018;&#x201D;(đ?&#x2018;Ľ)

lim

đ?&#x2018;&#x201C;(đ?&#x2018;Ľ)

đ?&#x2018;Ľâ&#x2020;&#x2019; â&#x2C6;&#x17E; đ?&#x2018;&#x201D;(đ?&#x2018;Ľ)

lim

đ?&#x2018;&#x201C;(đ?&#x2018;Ľ)

đ?&#x2018;Ľâ&#x2020;&#x2019; â&#x2C6;&#x17E; đ?&#x2018;&#x201D;(đ?&#x2018;Ľ)

=â&#x2C6;&#x17E;

=đ??ż

đ?&#x2018;&#x17D;

â&#x2C6;&#x2019;â&#x2C6;&#x17E;

đ?&#x2018;&#x201C;(đ?&#x2018;Ľ) > đ?&#x2018;&#x201D;(đ?&#x2018;Ľ) ,0 < đ??ż < â&#x2C6;&#x17E;

=0

đ?&#x2018;&#x17D;â&#x2C6;&#x17E; = {

VI.

3 1 =â&#x2C6;&#x2019; 2 2

â&#x2C6;&#x17E; 0

0 ={ â&#x2C6;&#x17E;

0â&#x2C6;&#x2019;

3 3 =â&#x2C6;&#x2019; 2 2

7đ?&#x2018;Ľ 5 +đ?&#x2018;Ľ 3 +1

đ?&#x2018;&#x2019;đ?&#x2018;Ľ: lim

đ?&#x2018;Ľâ&#x2020;&#x2019; â&#x2C6;&#x17E;

đ?&#x2018;&#x201C;(đ?&#x2018;Ľ) = đ?&#x2018;&#x201D;(đ?&#x2018;Ľ)

5đ?&#x2018;Ľ 4 â&#x2C6;&#x2019;1

đ?&#x2018;&#x2019;đ?&#x2018;Ľ: lim

7đ?&#x2018;Ľ 4 +đ?&#x2018;Ľ 3 +1

đ?&#x2018;&#x2019;đ?&#x2018;Ľ: lim

đ?&#x2018;&#x17D;>1 đ?&#x2018;&#x17D;<1

đ?&#x2018;&#x2019;đ?&#x2018;Ľ: 5â&#x2C6;&#x17E; = â&#x2C6;&#x17E;,

đ?&#x2018;&#x17D;>1 đ?&#x2018;&#x17D;<1

â&#x;š

7đ?&#x2018;Ľ 5 +đ?&#x2018;Ľ 3 +1 5đ?&#x2018;Ľ 5 â&#x2C6;&#x2019;1

đ?&#x2018;Ľâ&#x2020;&#x2019; â&#x2C6;&#x17E;

đ?&#x2018;&#x201C;(đ?&#x2018;Ľ) < đ?&#x2018;&#x201D;(đ?&#x2018;Ľ) â&#x;š

=â&#x2C6;&#x17E;

đ?&#x2018;Ľâ&#x2020;&#x2019; â&#x2C6;&#x17E;

đ?&#x2018;&#x2019;đ?&#x2018;Ľ: 5

â&#x2C6;&#x2019;â&#x2C6;&#x17E;

5đ?&#x2018;Ľ 5 â&#x2C6;&#x2019;1

= 0,

=0

(Large Degree Polynomial Function) =

7 5

(Equal Degree Polynomial Functions)

(Small Degree Polynomial Function)

1 â&#x2C6;&#x17E;

( ) =0 5

1 â&#x2C6;&#x2019;â&#x2C6;&#x17E; ( ) =â&#x2C6;&#x17E; 5

Section 1: Problems session đ??¸đ?&#x2018;Łđ?&#x2018;&#x17D;đ?&#x2018;&#x2122;đ?&#x2018;˘đ?&#x2018;&#x17D;đ?&#x2018;Ąđ?&#x2018;&#x2019; đ?&#x2018;Ąâ&#x201E;&#x17D;đ?&#x2018;&#x2019; đ?&#x2018;&#x201C;đ?&#x2018;&#x153;đ?&#x2018;&#x2122;đ?&#x2018;&#x2122;đ?&#x2018;&#x153;đ?&#x2018;¤đ?&#x2018;&#x2013;đ?&#x2018;&#x203A;đ?&#x2018;&#x201D; đ?&#x2018;&#x2122;đ?&#x2018;&#x2013;đ?&#x2018;&#x161;đ?&#x2018;&#x2013;đ?&#x2018;Ąđ?&#x2018; . đ??źđ?&#x2018;&#x201C; đ?&#x2018;Ąâ&#x201E;&#x17D;đ?&#x2018;&#x2019; đ?&#x2018;&#x2122;đ?&#x2018;&#x2013;đ?&#x2018;&#x161;đ?&#x2018;&#x2013;đ?&#x2018;Ą đ?&#x2018;&#x2018;đ?&#x2018;&#x153;đ?&#x2018;&#x2019;đ?&#x2018; đ?&#x2018;&#x203A;đ?&#x2018;&#x153;đ?&#x2018;Ą đ?&#x2018;&#x2019;đ?&#x2018;Ľđ?&#x2018;&#x2013;đ?&#x2018; đ?&#x2018;Ą, đ?&#x2018;&#x2013;đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;Ą â&#x2C6;&#x17E;, â&#x2C6;&#x2019;â&#x2C6;&#x17E; đ?&#x2018;&#x153;đ?&#x2018;&#x; đ?&#x2018;&#x203A;đ?&#x2018;&#x2019;đ?&#x2018;&#x2013;đ?&#x2018;Ąâ&#x201E;&#x17D;đ?&#x2018;&#x2019;đ?&#x2018;&#x;? (đ??ˇđ?&#x2018;&#x153; đ?&#x2018;&#x203A;đ?&#x2018;&#x153;đ?&#x2018;Ą đ?&#x2018;˘đ?&#x2018; đ?&#x2018;&#x2019; đ??żâ&#x20AC;˛ đ??ťĂ´đ?&#x2018;?đ?&#x2018;&#x2013;đ?&#x2018;Ąđ?&#x2018;&#x17D;đ?&#x2018;&#x2122; â&#x20AC;˛ đ?&#x2018; đ?&#x2018;&#x2026;đ?&#x2018;˘đ?&#x2018;&#x2122;đ?&#x2018;&#x2019;)

Part 1 (technique 1) 1.

2.

lim

đ?&#x2018;Ľâ&#x2020;&#x2019;1

lim

đ?&#x2018;Ľâ&#x2020;&#x2019;1

đ?&#x2018;Ľ 2 â&#x2C6;&#x2019;2đ?&#x2018;Ľ+1 đ?&#x2018;Ľâ&#x2C6;&#x2019;1

tan đ?&#x2018;Ľâ&#x2C6;&#x2019;đ?&#x2018;Ľ đ?&#x2018;Ľ3

Copyright Š 2019 Road To success All Rights Reserved. Email: hassanin.risha@gmail.com Website: https://math.gnomio.com

6

3.

4.

lim

6.

7.

8.

lim

đ?&#x2018;&#x2019; đ?&#x2018;Ľ +3đ?&#x2018;Ľ+1 sin đ?&#x2018;Ľâ&#x2C6;&#x2019;1

đ?&#x2018;&#x2019;đ?&#x2018;Ľ

đ?&#x2018;Ľâ&#x2020;&#x2019;â&#x2C6;&#x17E; sinh đ?&#x2018;Ľ

lim

đ?&#x2018;Ľâ&#x2020;&#x2019;3

lim

â&#x2020;? đ?&#x2018;&#x2026;đ?&#x2018;&#x2019;đ?&#x2018;Łđ?&#x2018;&#x2013;đ?&#x2018;&#x2019;đ?&#x2018;¤ đ?&#x2018;&#x2020;đ?&#x2018;&#x2019;đ?&#x2018;?đ?&#x2018;Ąđ?&#x2018;&#x2013;đ?&#x2018;&#x153;đ?&#x2018;&#x203A; 3

đ?&#x2018;Ľ+sin đ?&#x2018;Ľâ&#x2C6;&#x2019;cos đ?&#x2018;Ľ

lim đ?&#x153;&#x2039;â&#x2C6;&#x2019;

đ?&#x2018;Ľâ&#x2020;&#x2019; 2

5.

arcsin đ?&#x2018;Ľ +3 ln(đ?&#x2018;Ľ+1)

đ?&#x2018;Ľâ&#x2020;&#x2019;0

â&#x2020;? đ?&#x2018;&#x2026;đ?&#x2018;&#x2019;đ?&#x2018;Łđ?&#x2018;&#x2013;đ?&#x2018;&#x2019;đ?&#x2018;¤ đ?&#x2018;&#x2020;đ?&#x2018;&#x2019;đ?&#x2018;?đ?&#x2018;Ąđ?&#x2018;&#x2013;đ?&#x2018;&#x153;đ?&#x2018;&#x203A; 3

(đ?&#x2018;Ľ 2 +đ?&#x2018;Ľâ&#x2C6;&#x2019;12)â&#x2C6;&#x161;đ?&#x2018;Ľ+3 â&#x2C6;&#x161;5(đ?&#x2018;Ľ 2 â&#x2C6;&#x2019;9)

1â&#x2C6;&#x2019;cos đ?&#x2018;Ľ

đ?&#x2018;Ľâ&#x2020;&#x2019;0 sin2 đ?&#x2018;Ľ

lim đ?&#x2018;Ľ sin đ?&#x2018;Ľ

đ?&#x2018;Ľâ&#x2020;&#x2019;0

Part 2 (technique 2) 9.

lim

đ?&#x2018;Ľ 2 â&#x2C6;&#x2019;4

đ?&#x2018;Ľâ&#x2020;&#x2019;2 3đ?&#x2018;Ľ|đ?&#x2018;Ľâ&#x2C6;&#x2019;2|

Copyright Š 2019 Road To success All Rights Reserved. Email: hassanin.risha@gmail.com Website: https://math.gnomio.com

7

10.

11.

12.

13.

14.

2đ?&#x2018;Ľâ&#x2C6;&#x2019;2

lim |đ?&#x2018;Ľ 3

đ?&#x2018;Ľâ&#x2020;&#x2019;1

lim

â&#x2C6;&#x2019;đ?&#x2018;Ľ 2 |

|đ?&#x2018;Ľ+2|

đ?&#x2018;Ľâ&#x2020;&#x2019;â&#x2C6;&#x2019;2 |đ?&#x2018;Ľ|â&#x2C6;&#x2019;2

lim

|đ?&#x2018;Ľ 2 â&#x2C6;&#x2019;5|â&#x2C6;&#x2019;|đ?&#x2018;Ľ+3| đ?&#x2018;Ľâ&#x2C6;&#x2019;1

đ?&#x2018;Ľâ&#x2020;&#x2019;1

lim

|2đ?&#x2018;Ľâ&#x2C6;&#x2019;3|â&#x2C6;&#x2019;|đ?&#x2018;Ľâ&#x2C6;&#x2019;3| đ?&#x2018;Ľ

đ?&#x2018;Ľâ&#x2020;&#x2019;0

lim

đ?&#x2018;Ľâ&#x2020;&#x2019;0

|sin(đ?&#x2018;Ľ)| đ?&#x2018;Ľ

Copyright Š 2019 Road To success All Rights Reserved. Email: hassanin.risha@gmail.com Website: https://math.gnomio.com

8

15.

16.

lim

|đ?&#x2018;Ľ 2 â&#x2C6;&#x2019;9| 3

đ?&#x2018;Ľâ&#x2020;&#x2019;3 â&#x2C6;&#x161;đ?&#x2018;Ľâ&#x2C6;&#x2019;3

limđ?&#x153;&#x2039;

arctan(tan(đ?&#x2018;Ľ))

đ?&#x2018;Ľâ&#x2020;&#x2019; 2 tan(arctan(đ?&#x2018;Ľ))

â&#x2020;? đ?&#x2018;&#x2026;đ?&#x2018;&#x2019;đ?&#x2018;Łđ?&#x2018;&#x2013;đ?&#x2018;&#x2019;đ?&#x2018;¤ đ?&#x2018;&#x2020;đ?&#x2018;&#x2019;đ?&#x2018;?đ?&#x2018;Ąđ?&#x2018;&#x2013;đ?&#x2018;&#x153;đ?&#x2018;&#x203A; 3

Part 3 (technique 3) 17.

2sin(đ?&#x2018;Ľ 5 )

lim

đ?&#x2018;Ľâ&#x2020;&#x2019;â&#x2C6;&#x17E;

đ?&#x2018;Ľ 4 +1

1

18.

19.

lim

đ?&#x2018;Ľ 3 sin(đ?&#x2018;Ľ)

đ?&#x2018;Ľâ&#x2020;&#x2019;0 sin(đ?&#x2018;Ľ)

lim

đ?&#x2018;Ľâ&#x2020;&#x2019;â&#x2C6;&#x17E;

đ?&#x2018;?đ?&#x2018;&#x153;đ?&#x2018; (đ?&#x2018;Ľ sin đ?&#x2018;Ľ) đ?&#x2018;Ľ2

Copyright Š 2019 Road To success All Rights Reserved. Email: hassanin.risha@gmail.com Website: https://math.gnomio.com

9

20.

21.

22.

23.

24.

25.

lim

đ?&#x2018;Ľâ&#x2020;&#x2019;â&#x2C6;&#x17E;

lim

đ?&#x2018;Ľâ&#x2020;&#x2019;â&#x2C6;&#x17E;

lim

đ?&#x2018;Ľâ&#x2020;&#x2019;â&#x2C6;&#x17E;

lim

đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A; (đ?&#x2018;Ľ 2 +1) đ?&#x2018;Ľ2

đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A; (đ?&#x2018;Ľ 119 ) ln(đ?&#x2018;Ľ)

3đ?&#x2018;Ľ + |đ?&#x2018; đ?&#x2018;&#x2013;đ?&#x2018;&#x203A; (2đ?&#x2018;Ľ)| đ?&#x2018;Ľ2

đ?&#x2018;Ľ 2 +sin đ?&#x2018;Ľ

đ?&#x2018;Ľâ&#x2020;&#x2019;â&#x2C6;&#x17E; đ?&#x2018;Ľ 2 +sin2 đ?&#x2018;Ľ

lim

đ?&#x2018;Ľâ&#x2020;&#x2019;â&#x2C6;&#x17E;

lim

cos(â&#x2C6;&#x161;đ?&#x2018;Ľ 2 +1) đ?&#x2018;&#x2019; đ?&#x2018;Ľ +119

ln(1+sin2 đ?&#x2018;Ľ)

đ?&#x2018;Ľâ&#x2020;&#x2019;â&#x2C6;&#x17E; arctan đ?&#x2018;Ľ+đ?&#x2018;&#x2019; đ?&#x2018;Ľ

â&#x2020;? đ?&#x2018;&#x2026;đ?&#x2018;&#x2019;đ?&#x2018;Łđ?&#x2018;&#x2013;đ?&#x2018;&#x2019;đ?&#x2018;¤ đ?&#x2018;&#x2020;đ?&#x2018;&#x2019;đ?&#x2018;?đ?&#x2018;Ąđ?&#x2018;&#x2013;đ?&#x2018;&#x153;đ?&#x2018;&#x203A; 3

Copyright Š 2019 Road To success All Rights Reserved. Email: hassanin.risha@gmail.com Website: https://math.gnomio.com

10

đ?&#x153;&#x2039;

26.

lim đ?&#x2018;Ľ 2 đ?&#x2018;&#x2019; sin(đ?&#x2018;Ľ)

đ?&#x2018;Ľâ&#x2020;&#x2019;0

1

27.

lim+ đ?&#x2018;&#x2019; sin(đ?&#x2018;Ľ) ln(1 + đ?&#x2018;Ľ)

đ?&#x2018;Ľâ&#x2020;&#x2019;0

Part 4 (technique 4) 4

28.

29.

30.

31.

lim

đ?&#x2018;Ľâ&#x2020;&#x2019;â&#x2C6;&#x17E;

lim

đ?&#x2018;Ľ+ â&#x2C6;&#x161;16đ?&#x2018;Ľ 6 +đ?&#x2018;Ľ 5 â&#x2C6;&#x2019;3đ?&#x2018;Ľ+7 5đ?&#x2018;Ľ+8â&#x2C6;&#x2019;â&#x2C6;&#x161;đ?&#x2018;Ľ 3 +đ?&#x2018;Ľâ&#x2C6;&#x2019;2

đ?&#x2018;Ľ 3 +â&#x2C6;&#x161;đ?&#x2018;Ľ 6 +2đ?&#x2018;Ľ 5 +1

đ?&#x2018;Ľâ&#x2020;&#x2019;â&#x2C6;&#x17E; đ?&#x2018;Ľ 3 +2đ?&#x2018;Ľ 2 +đ?&#x2018;Ľ+1

lim

đ?&#x2018;Ľâ&#x2020;&#x2019;â&#x2C6;&#x17E;

lim

â&#x2C6;&#x161;đ?&#x2018;Ľ 4 +đ?&#x2018;Ľ+1 3

â&#x2C6;&#x161;đ?&#x2018;Ľ 6 +1

â&#x2C6;&#x161;đ?&#x2018;Ľ 2 +đ?&#x2018;&#x2019;

đ?&#x2018;Ľâ&#x2020;&#x2019;â&#x2C6;&#x17E; đ?&#x2018;Ľ 2 +7

Copyright Š 2019 Road To success All Rights Reserved. Email: hassanin.risha@gmail.com Website: https://math.gnomio.com

11

32.

lim [ln(đ?&#x2018;Ľ 2 + 1) â&#x2C6;&#x2019; ln(đ?&#x2018;Ľ 2 â&#x2C6;&#x2019; 1)]

đ?&#x2018;Ľâ&#x2020;&#x2019;â&#x2C6;&#x17E;

â&#x2020;? đ?&#x2018;&#x2026;đ?&#x2018;&#x2019;đ?&#x2018;Łđ?&#x2018;&#x2013;đ?&#x2018;&#x2019;đ?&#x2018;¤ đ?&#x2018;&#x2020;đ?&#x2018;&#x2019;đ?&#x2018;?đ?&#x2018;Ąđ?&#x2018;&#x2013;đ?&#x2018;&#x153;đ?&#x2018;&#x203A; 3

Part 5 (technique 4 and 5) 33.

34.

35.

36.

lim (â&#x2C6;&#x161;đ?&#x2018;Ľ + â&#x2C6;&#x161;đ?&#x2018;Ľ â&#x2C6;&#x2019; â&#x2C6;&#x161;đ?&#x2018;Ľ)

đ?&#x2018;Ľâ&#x2020;&#x2019;â&#x2C6;&#x17E;

lim (đ?&#x2018;Ľ â&#x2C6;&#x2019; â&#x2C6;&#x161;đ?&#x2018;Ľ 2 + 4đ?&#x2018;Ľ)

đ?&#x2018;Ľâ&#x2020;&#x2019;â&#x2C6;&#x2019;â&#x2C6;&#x17E;

lim (đ?&#x2018;Ľ + â&#x2C6;&#x161;đ?&#x2018;Ľ 2 â&#x2C6;&#x2019; đ?&#x2018;Ľ â&#x2C6;&#x2019; 4)

đ?&#x2018;Ľâ&#x2020;&#x2019;â&#x2C6;&#x2019;â&#x2C6;&#x17E;

lim đ?&#x2018;Ľ â&#x2C6;&#x2019; â&#x2C6;&#x161;đ?&#x2018;Ľ 2 + đ?&#x2018;Ľ + 2

đ?&#x2018;Ľâ&#x2020;&#x2019;â&#x2C6;&#x17E;

Copyright Š 2019 Road To success All Rights Reserved. Email: hassanin.risha@gmail.com Website: https://math.gnomio.com

12

37.

38.

39.

40.

41.

lim đ?&#x2018;Ľ â&#x2C6;&#x2019; â&#x2C6;&#x161;đ?&#x2018;Ľ 2 â&#x2C6;&#x2019; 4đ?&#x2018;Ľ + 1

đ?&#x2018;Ľâ&#x2020;&#x2019;â&#x2C6;&#x17E;

lim â&#x2C6;&#x161;đ?&#x2018;Ľ 2 + 4đ?&#x2018;Ľ â&#x2C6;&#x2019; â&#x2C6;&#x161;đ?&#x2018;Ľ 2 â&#x2C6;&#x2019; 4đ?&#x2018;Ľ

đ?&#x2018;Ľâ&#x2020;&#x2019;â&#x2C6;&#x17E;

lim â&#x2C6;&#x161;2đ?&#x2018;Ľ 2 â&#x2C6;&#x2019; đ?&#x2018;Ľ + 1 â&#x2C6;&#x2019; â&#x2C6;&#x161;đ?&#x2018;Ľ 2 â&#x2C6;&#x2019; đ?&#x2018;Ľ + 5

đ?&#x2018;Ľâ&#x2020;&#x2019;â&#x2C6;&#x2019;â&#x2C6;&#x17E;

lim â&#x2C6;&#x161;đ?&#x2018;Ľ 2 + đ?&#x2018;Ľ â&#x2C6;&#x2019; â&#x2C6;&#x161;đ?&#x2018;Ľ

đ?&#x2018;Ľâ&#x2020;&#x2019;â&#x2C6;&#x17E;

lim â&#x2C6;&#x161;đ?&#x2018;Ľ 10/3 â&#x2C6;&#x2019; 5đ?&#x2018;Ľ 2 â&#x2C6;&#x2019; â&#x2C6;&#x161;đ?&#x2018;Ľ 10/3 + 5đ?&#x2018;Ľ 2

đ?&#x2018;Ľâ&#x2020;&#x2019;â&#x2C6;&#x2019;â&#x2C6;&#x17E;

Copyright Š 2019 Road To success All Rights Reserved. Email: hassanin.risha@gmail.com Website: https://math.gnomio.com

13

42.

43.

44.

lim ln(â&#x2C6;&#x161;đ?&#x2018;Ľ 2 + 1 + đ?&#x2018;Ľ)

đ?&#x2018;Ľâ&#x2020;&#x2019;â&#x2C6;&#x2019;â&#x2C6;&#x17E;

lim

1â&#x2C6;&#x2019;cos đ?&#x2018;Ľ đ?&#x2018;Ľ2

đ?&#x2018;Ľâ&#x2020;&#x2019;0

lim

đ?&#x2018;Ľâ&#x2020;&#x2019;0

â&#x2C6;&#x161;1+tan đ?&#x2018;Ľâ&#x2C6;&#x2019;â&#x2C6;&#x161;1+sin đ?&#x2018;Ľ đ?&#x2018;Ľ3

Part 6 (technique 6) 45.

46.

47.

2â&#x2C6;&#x2019;đ?&#x2018;Ľ+sin(đ?&#x2018;Ľ)

lim

đ?&#x2018;Ľâ&#x2020;&#x2019;â&#x2C6;&#x17E; đ?&#x2018;Ľ+cos(đ?&#x2018;Ľ)

lim

2đ?&#x2018;Ľ 2 +sin2 đ?&#x2018;Ľ

đ?&#x2018;Ľâ&#x2020;&#x2019;â&#x2C6;&#x17E;

lim

đ?&#x2018;Ľâ&#x2020;&#x2019;0

đ?&#x2018;Ľ2

sin 5đ?&#x2018;Ľâ&#x2C6;&#x2019;sin 3đ?&#x2018;Ľ sin đ?&#x2018;Ľ

Copyright Š 2019 Road To success All Rights Reserved. Email: hassanin.risha@gmail.com Website: https://math.gnomio.com

14

48.

lim+

(tan â&#x2C6;&#x161;đ?&#x2018;Ľ )(â&#x2C6;&#x161;2đ?&#x2018;Ľ+1) đ?&#x2018;Ľ

đ?&#x2018;Ľâ&#x2020;&#x2019;0

Part 7 (technique 7)

â&#x2020;? đ?&#x2018;&#x2026;đ?&#x2018;&#x2019;đ?&#x2018;Łđ?&#x2018;&#x2013;đ?&#x2018;&#x2019;đ?&#x2018;¤ đ?&#x2018;&#x2020;đ?&#x2018;&#x2019;đ?&#x2018;?đ?&#x2018;Ąđ?&#x2018;&#x2013;đ?&#x2018;&#x153;đ?&#x2018;&#x203A; 3 đ?&#x153;&#x2039;

49.

lim1 đ?&#x2018;Ľâ&#x2020;&#x2019;2

50.

51.

lim

đ?&#x2018;Ľâ&#x2020;&#x2019;1

lim

đ?&#x2018;Ľâ&#x2020;&#x2019;0

arccos(đ?&#x2018;Ľ)â&#x2C6;&#x2019; 3 1

đ?&#x2018;Ľâ&#x2C6;&#x2019;2

arcsin(đ?&#x2018;Ľ 3 â&#x2C6;&#x2019;1) đ?&#x2018;Ľâ&#x2C6;&#x2019;1

cos đ?&#x2018;Ľ â&#x2C6;&#x2019; đ?&#x2018;&#x2019; đ?&#x2018;Ľ đ?&#x2018;Ľ

2

52.

53.

lim

đ?&#x2018;Ľâ&#x2020;&#x2019;0

lim

3(2+đ?&#x2018;Ľ) â&#x2C6;&#x2019;81 đ?&#x2018;Ľ

2đ?&#x2018;Ľ â&#x2C6;&#x2019; 1

đ?&#x2018;Ľâ&#x2020;&#x2019;0 sin 4đ?&#x2018;Ľ

Copyright Š 2019 Road To success All Rights Reserved. Email: hassanin.risha@gmail.com Website: https://math.gnomio.com

15

cos[(𝑥+2)2 ] −cos 4

lim

54.

𝑥

𝑥→0

Section 2: Continuity and Differentiation Continuity: 𝑅𝑢𝑙𝑒: 𝐼𝑓 𝑓(𝑥) 𝑖𝑠 𝑐𝑜𝑛𝑡𝑖𝑛𝑜𝑢𝑠 𝑎𝑡 𝑥 = 𝑎 ⇄

𝑇ℎ𝑒𝑛,

lim 𝑓(𝑥) = 𝑓(𝑎)

𝑥→ 𝑎

𝑂𝑅

lim 𝑓(𝑥) = lim− 𝑓(𝑥) = 𝑓(𝑎)

𝑥→ 𝑎+

𝑥→ 𝑎

Differentiation: (Rules) (Formal) Definition of the Derivative 𝑓(𝑥 + ℎ) − 𝑓(𝑥) ℎ→ 0 ℎ 𝐼𝑓 𝑓(𝑥) 𝑖𝑠 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑏𝑙𝑒 𝑎𝑡 𝑥 = 𝑎 ⇄

Definition of the Derivative 𝑓(𝑥) − 𝑓(𝑎) 𝑥→ 𝑎 𝑥−𝑎 𝐼𝑓 𝑓(𝑥) 𝑖𝑠 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑏𝑙𝑒 𝑎𝑡 𝑥 = 𝑎

𝑓′(𝑥) = lim

𝑓′(𝑎) = lim

𝑇ℎ𝑒𝑛,

𝑓(𝑥 + ℎ) − 𝑓(𝑥) 𝑓(𝑥 + ℎ) − 𝑓(𝑥) = lim− ℎ→ 0 ℎ→ 0 ℎ ℎ 𝑓(𝑥 + ℎ) − 𝑓(𝑥) 𝑂𝑅 𝑓′(𝑥) = lim ℎ→ 0 ℎ

𝑓 ′ (𝑥) = lim+

𝑓 ′ (𝑎) = lim+ 𝑥→ 𝑎

𝑂𝑅

⇄

𝑇ℎ𝑒𝑛,

𝑓(𝑥) − 𝑓(𝑎) 𝑓(𝑥) − 𝑓(𝑎) = lim+ 𝑥→ 𝑎 𝑥−𝑎 𝑥−𝑎 𝑓(𝑥) − 𝑓(𝑎) 𝑓′(𝑎) = lim 𝑥→ 𝑎 𝑥−𝑎

𝑰𝒎𝒑𝒐𝒓𝒕𝒂𝒏𝒕: 𝐼𝑓 𝑓(𝑥) 𝑖𝑠 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑏𝑙𝑒 𝑎𝑡 𝑥 = 𝑎 𝑇ℎ𝑒𝑛, 𝑓(𝑥) 𝑖𝑠 𝑐𝑜𝑛𝑡𝑖𝑛𝑜𝑢𝑠 𝑎𝑡 𝑥 = 𝑎 (NOT the inverse) Examples: −𝑥, 𝑥 < 0 𝑥2, 𝑥 ≥ 0

1- Example: 𝑓(𝑥) = {

Answer: ∵ lim+ 𝑥 2 = 𝑓(0) = 0 𝑥→ 0

−𝑥, 𝑥 ≠ 0 𝑥2, 𝑥 = 0

2- Example: 𝑓(𝑥) = { Answer: ∵ lim 𝑥 2 = 0

𝑎𝑛𝑑

lim −𝑥 = 0

𝑥→ 0−

−𝑥, 𝑥 < 0 𝑥2, 𝑥 ≥ 0

∴ 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑓 𝑖𝑠 𝑐𝑜𝑛𝑡𝑖𝑛𝑜𝑢𝑠 𝑎𝑡 𝑥 = 0

𝐹𝑖𝑛𝑑 𝑓′(0) 𝑖𝑓 𝑖𝑡 𝑒𝑥𝑖𝑠𝑡𝑠

Answer using (Formal) Definition of the Derivative 𝑓(𝑥 + ℎ) − 𝑓(𝑥) 𝑓′(𝑥) = lim ℎ→ 0 ℎ 𝑓(0 + ℎ) − 𝑓(0) 𝑓(ℎ) 𝑓′(0) = lim = lim ℎ→ 0 ℎ→ 0 ℎ ℎ 𝑓(ℎ) ℎ2 lim+ = lim+ = ℎ = 0 , ℎ→ 0

lim

ℎ 𝑓(ℎ)

ℎ→ 0

= lim

ℎ −ℎ

=−1

ℎ→ 0− ℎ ℎ→ 0− ℎ 𝑓(ℎ) 𝑓(ℎ)

∵ lim+ ℎ→ 0

ℎ

≠ lim− ℎ→ 0

∴ 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑓 𝑖𝑠 𝑐𝑜𝑛𝑡𝑖𝑛𝑜𝑢𝑠 𝑎𝑡 𝑥 = 0

𝑠ℎ𝑜𝑤 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑓 𝑖𝑠 𝑐𝑜𝑛𝑡𝑖𝑛𝑜𝑢𝑠 𝑎𝑡 𝑥 = 0

𝑎𝑛𝑑 𝑓(0) = 0

𝑥→ 0

3- Example: 𝑓(𝑥) = {

𝑠ℎ𝑜𝑤 𝑡ℎ𝑎𝑡 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑓 𝑖𝑠 𝑐𝑜𝑛𝑡𝑖𝑛𝑜𝑢𝑠 𝑎𝑡 𝑥 = 0

ℎ

∴ 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑓 𝑖𝑠 𝑁𝑂𝑇 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑏𝑙𝑒 𝑎𝑡 𝑥 = 0

Answer using Definition of the Derivative 𝑓(𝑥) − 𝑓(𝑎) 𝑓′(𝑎) = lim 𝑥→ 𝑎 𝑥−𝑎 𝑓(𝑥) − 𝑓(0) 𝑓(𝑥) 𝑓′(0) = lim = lim 𝑥→ 0 𝑥→ 0 𝑥 𝑥−0 𝑓(𝑥) 𝑥2 lim+ = lim+ = 𝑥 = 0 , 𝑥→ 0

lim

𝑥 𝑓(𝑥)

𝑥→ 0

= lim

𝑥 −𝑥

=−1

𝑥→ 0− 𝑥 𝑥→ 0− 𝑥 𝑓(𝑥) 𝑓(𝑥)

∵ lim+ 𝑥→ 0

𝑥

≠ lim− 𝑥→ 0

𝑥

∴ 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 𝑓 𝑖𝑠 𝑁𝑂𝑇 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑡𝑖𝑎𝑏𝑙𝑒 𝑎𝑡 𝑥 = 0

Copyright © 2019 Road To success All Rights Reserved. Email: hassanin.risha@gmail.com Website: https://math.gnomio.com

16