FSKM/ UiTM Pahang © Amirah Hana
FSKM/ UiTM Pahang © Amirah Hana Limits and Continuity Differentiation Applications of Differentiation Integration Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Applications of Integration
FSKM/ UiTM Pahang © Amirah Hana
FSKM/ UiTM Pahang © Amirah Hana At the end of this session, the students should be able to ❶ Find algebraically the limits of functions ❷ Recognize and determine infinite limits and limits at infinity. ❸ Determine the existence of the limits of functions. ➍ Determine whether the function is continuous or discontinuous at =
FSKM/ UiTM Pahang © Amirah Hana Limits are all about approaching. Sometimes you can't work something out directly, but you can see what it should be as you get closer and closer! For example, given = 2 − 4 − 2 Let’s work it out for = 2 2 = 2 2 − 4 2 − 2 = 0 0 Now 0 0 is a difficulty! We don't really know the value of 0 0 (it is "indeterminate"), so we need another way of answering this. calculator will give error when we calculate 0 0 !
FSKM/ UiTM Pahang © Amirah Hana x = 2 − 4 − 2 1.5 3.5 1.9 3.9 1.99 3.99 1.999 3.999 1.9999 3.9999 1.99999 3.99999 1.999999 3.999999 … … So instead of trying to work it out for x=2 let’s try approaching it closer and closer: as x gets close to 2 then 2−4 −2 gets close to 4 We are now faced with an interesting situation: When x=2 we don't know the answer (it is indeterminate) But we can see that if as x gets close to 2, it is going to be 4 2 = 2 2 − 4 2 − 2 = 0 0 =
FSKM/ UiTM Pahang © Amirah Hana We want to give the answer “4" but can't, so instead mathematicians say exactly what is going on by using the special word "limit". So, it is a special way of saying, "ignoring what happens when we get there, but as we get x closer and closer to 2, the answer gets closer and closer to 4" And it is written in symbols as: The limit of 2−4 −2 as x approaches 2 is 4 lim →2 2 − 4 − 2 = 4
FSKM/ UiTM Pahang © Amirah Hana Limits can be used even when we know the value when we get there! Nobody said they are only for difficult functions. For example, lim →5 2 = 10 We know perfectly well that 2(5) = 10, but limits can still be used (if we want!)
FSKM/ UiTM Pahang © Amirah Hana 1. Constant Rule 2. Constant Multiple Rule 3. Addition & Subtraction Rule 4. Multiplication Rule 5. Division Rule 6. Power Rule 7. Root Rule lim → = lim → ± = lim → ± lim → lim → = lim → lim → lim → = lim → lim → = lim → lim → ∙ = lim → ∙ lim → lim → = lim → lim → ≠ 0 n +ve integer
FSKM/ UiTM Pahang © Amirah Hana there are several ways to evaluate the limits ➊ Evaluating Limits using Direct Substitution ➋ Evaluating Limits using Factoring and Canceling ➌ Evaluating Limits using Multiplying by the Conjugate ➍ Evaluating Limits using Squeezing Theorem
FSKM/ UiTM Pahang © Amirah Hana lim −2 →1 a) −2 = lim 3 →0 b) 3 = lim →−2 c) = The limit of a constant is the constant There is no variable x to substitute!
FSKM/ UiTM Pahang © Amirah Hana lim 2 →1 2 3 = 2 lim →1 3 lim →1 2 3 = Theorem : Bring out constant from ‘lim’ expression 2 1 3 = = 2 1 = 2 3 After direct substitution, no more ‘lim’ expression
FSKM/ UiTM Pahang © Amirah Hana lim 3 →2 3 − 3 + 1 = lim →2 3 − lim →2 3 + lim →2 lim 1 →2 3 − 3 + 1 = Theorem : separating limits 2 3 − 3 2 + 1 = 8 − 6 + 1 = = 3 = 2 3 − 3 2 + 1 = 8 − 6 + 1 After direct substitution, no more ‘lim’ expression
FSKM/ UiTM Pahang © Amirah Hana lim −2 →−1 2 − 1 = lim →−1 2 ∙ lim →−1 lim − 1 →−1 2 − 1 = Theorem : separating limits −1 2 −1 − 1 = 1 −2 = = −2 = −1 2 −1 − 1 = 1 −2 After direct substitution, no more ‘lim’ expression
FSKM/ UiTM Pahang © Amirah Hana lim −1 →0 − 1 + 1 = lim →0 ( − 1) lim →0 ( + 1) lim →0 − 1 + 1 = 0 − 1 0 + 1 = −1 1 = = = −1 0 − 1 0 + 1 = −1 1 After direct substitution, no more ‘lim’ expression lim → ≠ 0 Theorem : separating the division of limits
FSKM/ UiTM Pahang © Amirah Hana lim →2 3 = lim →2 3 lim →2 3 = Theorem : The limit of the cube of a function equals the cube of the limit of a function 2 3 = 8 = 2 3 = 8 After direct substitution, no more ‘lim’ expression move out the power of n, from limits expression
FSKM/ UiTM Pahang © Amirah Hana lim →2 3 4 = 3 lim →2 lim 4 →2 3 4 = Theorem : The limit of the cube root of a function equals the cube root of the limit of a function 3 4 2 = 3 8 = = 3 4 2 = 3 8 move out the nroot, from limits expression 2 = 2 After direct substitution, no more ‘lim’ expression
Functions with Direct Substitution Property are called continuous at a. FSKM/ UiTM Pahang © Amirah Hana However, not all limits can be evaluated by direct substitution. The next video are some other techniques that can be used. As long as the answer gives a real number, we can use direct substitution to evaluate limits.
FSKM/ UiTM Pahang © Amirah Hana there are several ways to evaluate the limits ➊ Evaluating Limits using Direct Substitution ➋ Evaluating Limits using Factoring and Canceling ➌ Evaluating Limits using Multiplying by the Conjugate ➍ Evaluating Limits using Squeezing Theorem → =
FSKM/ UiTM Pahang © Amirah Hana lim → = 0 lim → = 0 lim → ≠ 0 lim → = lim → lim → = 0 0 lim → = 0 0
FSKM/ UiTM Pahang © Amirah Hana Did you know that 0 0 ≠ 1 0 0 is undefined ! Calculator gives ‘ERROR’ lim → = 0 0
FSKM/ UiTM Pahang © Amirah Hana In this case, we can’t evaluate limits by direct substitution. lim → = 0 0 Using direct substitution, if we get… lim → = 0 0
FSKM/ UiTM Pahang © Amirah Hana The following are some other techniques that can be used. lim → = 0 0
FSKM/ UiTM Pahang © Amirah Hana lim → = 0 0 lim → = 0 0 When or is a quadratic & polynomial Fractions within fractions There is no term containing “surd’ or “trigonometric” function
FSKM/ UiTM Pahang © Amirah Hana Let’s look at some examples… lim → = 0 0
FSKM/ UiTM Pahang © Amirah Hana lim →2 2 − 4 − 2 Find Using direct substitution, lim →2 2 − 4 − 2 = 2 2 − 4 2 − 2 = 0 0 Error! lim →2 2 − 4 − 2 = lim →2 + 2 − 2 − 2 factoring = lim →2 + 2 canceling = 2 + 2 After direct substitution, no more ‘lim’ expression = 4 #
FSKM/ UiTM Pahang © Amirah Hana lim →1 1 − 2 − 1 Find Using direct substitution, lim →1 1 − 2 − 1 = 1 − 1 2 1 − 1 = 0 0 Error! lim →1 1 − 2 − 1 = lim →1 1 + 1 − − 1 factoring = −lim →1 1 + canceling = − 1 + 1 After direct substitution, no more ‘lim’ expression = −2 # Are you sure can canceling?? − ≠ − So, we can’t canceling! 1 − = − − 1 = − + 1 = lim →1 − ( − 1) 1 + − 1 canceling
FSKM/ UiTM Pahang © Amirah Hana As long as the answer gives a real number, we can use direct substitution to evaluate limits. However, not all limits can be evaluated by direct substitution. lim → = 0 0
FSKM/ UiTM Pahang © Amirah Hana lim →−1 3 + 2 2 + − Find 3 Using direct substitution, lim →−1 3 + 2 2 + − 3 = −1 3 + 2 −1 2 + (−1) −1 − −1 3 = 0 0 = −1 + 2 1 − 1 −1 − −1 = −1 + 2 − 1 −1 + 1 indeterminate Error! In this case, we can’t evaluate limits by direct substitution.
FSKM/ UiTM Pahang © Amirah Hana lim →−1 3 + 2 2 + − Find 3 lim →−1 3 + 2 2 + − 3 = lim →−1 2 + 2 + 1 1 − 2 factorize still 0/0 after direct substitution?? = lim →−1 + 1 + 1 1 + 1 − canceling = lim →−1 + 1 1 − = lim →−1 2 + 2 + 1 1 − 2 Since after direct substitution, we get 0/0... Factorize Again ! canceling Using direct substitution, lim →−1 2 + 2 + 1 1 − 2 = 0 0 = −1 2 + 2 −1 + 1 1 − −1 2 = 1 − 2 + 1 1 − 1 Error!
Since we factorize twice, we also can factorize like this FSKM/ UiTM Pahang © Amirah Hana lim →−1 3 + 2 2 + − Find 3 lim →−1 3 + 2 2 + − 3 = lim →−1 2 + 2 + 1 1 − 2 = 0 # = lim →−1 + 1 + 1 1 + 1 − = lim →−1 + 1 1 − = lim →−1 2 + 2 + 1 1 − 2 = −1 + 1 1 − −1 = −1 + 1 1 + 1 = 0 2 lim →−1 3 + 2 2 + − 3 = lim →−1 + 1 + 1 1 + 1 − = lim →−1 + 1 1 − After direct substitution, no more ‘lim’ expression
FSKM/ UiTM Pahang © Amirah Hana lim →−4 1 + 1 4 + 4 Find Using direct substitution, lim →−4 1 + 1 4 + 4 = 1 −4 + 1 4 −4 + 4 = 0 0 Error! lim →−4 1 + 1 4 + 4 = lim →−4 4 + 4 + 4 = lim →−4 4 + 4∙ 1 + 4 canceling = 1 4 −4 = − 1 16 # If there are fractions within fractions, first, try to combine the fractions. = lim →−4 1 4 lim →−4 4 + 4 + 4 = lim →−4 4 + 4 ÷ + 4 = lim →−4 + 4 4 ÷ + 4 1 = lim →−4 + 4 4∙ 1 + 4
= lim →0 − 1 1 − FSKM/ UiTM Pahang © Amirah Hana lim →0 2− 1 − Find Using direct substitution, lim →0 2− 1 − = 0 − 0 1 − 0 = 0 0 Error! lim →0 2− 1 − = lim →0 ∙ − 1 − = 1 − 1 1 − 1 madam, why 2 = ∙ ? From the properties of indices : ∙ = + = 2 ∙ = + canceling Are you sure can canceling??
= lim →0 − 1 1 − FSKM/ UiTM Pahang © Amirah Hana lim →0 2− 1 − Find lim →0 2− 1 − = lim →0 ∙ − 1 − − ≠ − So, we can’t canceling! − 1 = − 1 − = −1 + = lim →0 − 1 − 1 −
= lim →0 − 1 1 − FSKM/ UiTM Pahang © Amirah Hana lim →0 2− 1 − Find lim →0 2− 1 − = lim →0 ∙ − 1 − = lim →0 − 1 − 1 − canceling = − lim →0 = − 0 After direct substitution, no more ‘lim’ expression 1 = −1 # The number e is one of the most important numbers in mathematics. The first few digits are : 2.7182818284590452353602874713527 (and more ...) e is an irrational number (it cannot be written as a simple fraction). e is the base of the Natural Logarithms (invented by John Napier). e is found in many interesting areas, so is worth learning about. It is often called Euler’s Number after the Swiss mathematician Leonhard Euler (pronounced "Oiler"). e (Euler’s Number) e
FSKM/ UiTM Pahang © Amirah Hana there are several ways to evaluate the limits ➊ Evaluating Limits using Direct Substitution ➋ Evaluating Limits using Factoring and Canceling ➌ Evaluating Limits using Multiplying by the Conjugate ➍ Evaluating Limits using Squeezing Theorem → =
FSKM/ UiTM Pahang © Amirah Hana The conjugate is where we change the sign in the middle of two terms. lim → = 0 0 We only use it in expressions with two terms, called "binomials"
FSKM/ UiTM Pahang © Amirah Hana lim → = 0 0 Binomial Sign change conjugate + 2 + to − − 2 3 − − to + 3 + 1 − + 2 + to − 1 − − 2 3 − − 1 − to + 3 + − 1 Radical expression that involve a sum or difference of the two terms are called conjugates
FSKM/ UiTM Pahang © Amirah Hana To make the expression free from radical sign! lim → = 0 0
FSKM/ UiTM Pahang © Amirah Hana Let’s consider this… + − = 2− + − 2 = 2− 2 + − = 2 − 2
FSKM/ UiTM Pahang © Amirah Hana Let’s try some examples… + − = 2 − 2 a) + 2 − 2 = 2 − 2 2 = − 4 No more radical sign! The products of two conjugates is the difference of two squares This result is very helpful when simplifying radical expressions
FSKM/ UiTM Pahang © Amirah Hana Let’s look at other examples… + − = 2 − 2 b) 3 − 3 + = 3 2 − 2 = 9 − No more radical sign!
FSKM/ UiTM Pahang © Amirah Hana How about… + − = 2 − 2 c) 1 − + 2 1 − − 2 = 1 − 2 − 2 2 = 1 − − 4 No more = −3 − radical sign!
FSKM/ UiTM Pahang © Amirah Hana Now, look at this examples… + − = 2 − 2 d) 3 − − 1 3 + − 1 = 3 2 − − 1 2 = 9 − ( − 1) = 9 − + 1 = 10 − Must put bracket ! No more radical sign!
FSKM/ UiTM Pahang © Amirah Hana lim →4 − 2 − 4 Find Using direct substitution, lim →4 − 2 − 4 = 4 − 2 4 − 4 = 0 0 Error! lim →4 − 2 − 4 = lim →4 − 2 − 4 ∙ + 2 + 2 = lim →4 − 2 + 2 − 4 + 2 Multiplying top and bottom by the conjugate conjugate each other Simplify using : + − = 2 − 2 Not conjugate each other No need to expand = 2 − 2 4 − 4 = lim →4 ( ) 2−(2) 2 − 4 + 2 1
FSKM/ UiTM Pahang © Amirah Hana lim →4 − 2 − 4 Find lim →4 − 2 − 4 = lim →4 − 2 − 4 ∙ + 2 + 2 = lim →4 − 2 + 2 − 4 + 2 = lim →4 ( ) 2−(2) 2 − 4 + 2 = lim →4 − 4 − 4 + 2 simplify ! = lim →4 1 + 2 = 1 2 + 2 After direct substitution, no more ‘lim’ expression = 1 4 + 2 = 1 4 #
FSKM/ UiTM Pahang © Amirah Hana lim →−1 + 1 + 2 − 1 Find Using direct substitution, lim →−1 + 1 + 2 − 1 = −1 + 1 −1 + 2 − 1 = 0 0 Error! = −1 + 1 1 − 1
FSKM/ UiTM Pahang © Amirah Hana lim →−1 + 1 + 2 − 1 = lim →−1 + 1 + 2 − 1 ∙ + 2 + 1 + 2 + 1 = lim →−1 + 1 + 2 + 1 + 2 − 1 + 2 + 1 Multiplying top and bottom by the conjugate = lim →−1 + 1 + 2 + 1 ( + 2) 2−(1) 2 1 conjugate each other Simplify using : + − = 2 − 2 Not conjugate each other No need to expand lim →−1 + 1 + 2 − 1 Find
FSKM/ UiTM Pahang © Amirah Hana lim →−1 + 1 + 2 − 1 = lim →−1 + 1 + 2 − 1 ∙ + 2 + 1 + 2 + 1 = lim →−1 + 1 + 2 + 1 + 2 − 1 + 2 + 1 = lim →−1 + 1 + 2 + 1 ( + 2) 2−(1) 2 lim →−1 + 1 + 2 − 1 Find = lim →−1 + 1 + 2 + 1 + 2 − 1 = lim →−1 + 1 + 2 + 1 + 1 simplify = lim →−1 + 2 + 1
FSKM/ UiTM Pahang © Amirah Hana lim →−1 + 1 + 2 − 1 = lim →−1 + 2 + 1 = −1 + 2 + 1 substitute After direct substitution, no more ‘lim’ expression = 1 + 1 = 2 # lim →−1 + 1 + 2 − 1 Find