FSKM/ UiTM Pahang © Amirah Hana Using direct substitution, lim →4 1 2 − 1 − 4 = 1 2 − 1 4 4 − 4 = 0 0 Error! = 1 2 − 1 2 4 − 4 lim →4 1 2 − 1 − 4 Find
FSKM/ UiTM Pahang © Amirah Hana lim→4 12 − 1 − 4 Find lim→4 12 − 1 − 4 = lim→4 − 2 2 − 4 = lim→4 − 2 2 ∙ 1 − 4 = lim→4 − 2 2 − 4 If there are fractions within fractions, first, try to combine the fractions. lim→4 − 2 2 − 4 = lim→4 − 2 2 ÷ − 4 = lim→4 − 2 2 ÷ − 4 1 = lim→4 − 2 2 ∙ 1 − 4 = lim→4 − 2 2 − 4
= lim →4 − 2 2 − 4 FSKM/ UiTM Pahang © Amirah Hana lim →4 1 2 − 1 − 4 Find lim →4 1 2 − 1 − 4 = lim →4 − 2 2 − 4 = lim →4 − 2 2 ∙ 1 − 4 conjugate is a radical expression that involve a sum or difference of the two terms
FSKM/ UiTM Pahang © Amirah Hana lim →4 1 2 − 1 − 4 Find lim →4 1 2 − 1 − 4 = lim →4 − 2 2 − 4 = lim →4 − 2 2 ∙ 1 − 4 = lim →4 − 2 2 − 4 = lim →4 − 2 2 − 4 ∙ + 2 + 2 Multiplying top and bottom by the conjugate 1 = lim →4 − 2 + 2 2 − 4 + 2 conjugate each other Simplify using : + − = 2 − 2 Not conjugate each other No need to expand = lim →4 ( ) 2−(2) 2 2 − 4 + 2
FSKM/ UiTM Pahang © Amirah Hana substitute = 1 2 4 4 + 2 After direct substitution, no more ‘lim’ expression = 1 2 2 4 = 1 16 # lim →4 1 2 − 1 − 4 Find lim →4 1 2 − 1 − 4 = lim →4 ( ) 2−(2) 2 2 − 4 + 2 = lim →4 − 4 2 − 4 + 2 simplify = lim →4 1 2 + 2
FSKM/ UiTM Pahang © Amirah Hana lim →1 2 + 3 − 2 − 1 Find Using direct substitution, lim →1 2 + 3 − 2 − 1 = 1 2 + 3 − 2 1 − 1 = 0 0 Error! = 2 − 2 1 − 1
FSKM/ UiTM Pahang © Amirah Hana lim →1 2 + 3 − 2 − 1 = lim →1 2 + 3 − 2 − 1 ∙ 2 + 3 + 2 2 + 3 + 2 = lim →1 2 + 3 − 2 2 + 3 + 2 − 1 2 + 3 + 2 Multiplying top and bottom by the conjugate = lim →1 ( 2 + 3) 2−(2) 2 − 1 2 + 3 + 2 conjugate each other Simplify using : + − = 2 − 2 lim →1 2 + 3 − 2 − 1 Find Not conjugate each other No need to expand
FSKM/ UiTM Pahang © Amirah Hana lim →1 2 + 3 − 2 − 1 = lim →1 2 + 3 − 2 − 1 ∙ 2 + 3 + 2 2 + 3 + 2 = lim →1 2 + 3 − 2 2 + 3 + 2 − 1 2 + 3 + 2 = lim →1 ( 2 + 3) 2−(2) 2 − 1 2 + 3 + 2 lim →1 2 + 3 − 2 − 1 Find = lim →1 2 + 3 − 4 − 1 2 + 3 + 2 = lim →1 2 − 1 − 1 2 + 3 + 2 we can simplify this… we need to factorize it first ! = lim →1 + 1 − 1 − 1 2 + 3 + 2
= lim →1 + 1 − 1 − 1 2 + 3 + 2 FSKM/ UiTM Pahang © Amirah Hana lim →1 2 + 3 − 2 − 1 lim →1 2 + 3 − 2 − 1 Find = lim →1 + 1 2 + 3 + 2 simplify substitute = 1 + 1 1 2 + 3 + 2 = 2 2 + 2 = 2 4 = 1 2 # After direct substitution, no more ‘lim’ expression
FSKM/ UiTM Pahang © Amirah Hana There are two special limits involving sine and cosine …
FSKM/ UiTM Pahang © Amirah Hana Let’s we look at this… The denominator and the argument of sine must be exact same expression in order to properly use the special trigonometric limit !
FSKM/ UiTM Pahang © Amirah Hana →0 Find 2 Using direct substitution, →0 2 = 2 0 = 0 accepted! = 0 defined No need to use Squeezing Theorem!
FSKM/ UiTM Pahang © Amirah Hana →0 2 Find Using direct substitution, →0 2 = 2 0 0 = 0 0 Error! = 0 0 In this case, we need to use Squeezing Theorem!
FSKM/ UiTM Pahang © Amirah Hana →0 2 Find →0 2 = →0 2 2 2 = 2 →0 2 2 → = = 2 1 = 2 #
FSKM/ UiTM Pahang © Amirah Hana →0 2 Find →0 2 = 2 →0 2 2 → = = 2 1 = 2 # →0 2 = →0 2 = 2 1 = 2 # → = = 2 →0 = →0 2 2 2
FSKM/ UiTM Pahang © Amirah Hana →0 2 Find →0 2 = 2 →0 2 2 → = = 2 1 = 2 # →0 2 = →0 2 = 2 1 = 2 # → = Can we use this technique?? = 2 →0 = →0 2 2 2
FSKM/ UiTM Pahang © Amirah Hana →0 2 Find →0 2 = 2 →0 2 2 → = = 2 1 = 2 # →0 2 = →0 2 = 2 1 = 2 # = 2 →0 Are you sure 2 = 2 ?? ≠ If = 30° 2 30° ≠ 2 30° 60° ≠ 2 0.5 0.866 ≠ 1 = →0 2 2 2 → =
FSKM/ UiTM Pahang © Amirah Hana →0 2 Find →0 2 = 2 →0 2 2 → = = 2 1 = 2 # →0 2 = →0 2 = 2 1 = 2 # = 2 →0 ≠ = →0 2 2 2
FSKM/ UiTM Pahang © Amirah Hana →0 2 Find →0 2 = 2 →0 2 2 → = = 2 1 = 2 # = →0 2 2 2
FSKM/ UiTM Pahang © Amirah Hana →0 5 2 Find Using direct substitution, →0 5 2 = 5 0 2 0 = 0 0 Error! = 0 0 In this case, we need to use Squeezing Theorem!
FSKM/ UiTM Pahang © Amirah Hana →0 52 Find →0 5 2 = 12 →0 5 5 5 → = = 52 1 = 52 # = 12 →0 5 = 52 →0 5 5
= 5 2 →0 5 5 FSKM/ UiTM Pahang © Amirah Hana →0 5 2 Find →0 5 2 = 1 2 →0 5 5 5 → = = 5 2 1 = 5 2 # = 1 2 →0 5 →0 5 2 = 5 2 →0 = 5 2 1 = 5 2 #
FSKM/ UiTM Pahang © Amirah Hana →0 3 4 Find Using direct substitution, →0 3 4 = 3 0 4 0 = 0 0 Error! = 0 0 In this case, we need to use Squeezing Theorem!
→0 3 4 Find FSKM/ UiTM Pahang © Amirah Hana →0 3 4 → = = →0 3 4∙ 1 1 → = = →0 3 4 = →0 3 →0 4
= 3 →0 3 3 4 →0 4 4 →0 3 4 Find FSKM/ UiTM Pahang © Amirah Hana →0 3 4 → = = →0 3 4∙ 1 1 → = = →0 3 4 = →0 3 3 3 →0 4 4 4 = 3 1 4 1 = 3 4 = # →0 3 →0 4
FSKM/ UiTM Pahang © Amirah Hana →0 Find 2 3 Using direct substitution, →0 2 3 = 0 0 Error! In this case, we need to use Squeezing Theorem! = 2 0 3 0 = 0 0 There is no ‘csc’ button on the calculator! = = 1 0 ∙ 0 = 0 0
→0 Find 2 3 FSKM/ UiTM Pahang © Amirah Hana →0 2 3 = →0 3 2∙ 1 3 = →0 1 2∙ 3 3 → = → =
→0 Find 2 3 FSKM/ UiTM Pahang © Amirah Hana →0 2 3 = →0 3 2∙ 1 3 = →0 1 2∙ 3 3 → = → =
→0 Find 2 3 FSKM/ UiTM Pahang © Amirah Hana →0 2 3 = →0 3 2∙ 1 3 = →0 1 2∙ 3 3 → = → = 1 5 ∙ 3 2 = 3 10 3 5 ∙ 1 2 = 3 10
→0 Find 2 3 FSKM/ UiTM Pahang © Amirah Hana →0 2 3 = →0 3 2∙ 1 3 = →0 1 2∙ 3 3 → = → = = →0 3 2∙ →0 1 3 Squeezing theorem Direct substitution = = =
→0 Find 2 3 FSKM/ UiTM Pahang © Amirah Hana →0 2 3 = →0 3 2∙ 1 3 = →0 1 2∙ 3 3 → = → = = →0 3 2∙ →0 1 3 = →0 3 2∙ 1 1 ∙ 1 0 = →0 3 2 ∙ 1 1 = →0 3 2 = →0 3 →0 2
→0 Find 2 3 FSKM/ UiTM Pahang © Amirah Hana →0 2 3 = →0 3 →0 2 → = → = = →0 3 33 →0 2 22 = 3 →0 33 2 →0 22 = 3 1 2 1 = 32 #
FSKM/ UiTM Pahang © Amirah Hana →0 3 2 − 2 2 Find 2 Using direct substitution, →0 3 2 − 2 2 2 = 0 0 Error! In this case, we need to use Squeezing Theorem! = 3 0 2 − 2 0 2 0 2 = 0 − 0 2 0
→0 3 2 − 2 2 Find 2 FSKM/ UiTM Pahang © Amirah Hana →0 3 2 − 2 2 2 = →0 3 2 − 2 2 2 = →0 3 2 2 2 − 2 2 2 → = = →0 3 2 − →0 2 2 2 = 3 2 − →0 1 2 ∙ ∙ = 3 2 − →0 1 2 ∙ →0 ∙ →0 = 3 2 − 1 2 ∙ 1 ∙ 1 = 3 2 − 1 2 = 2 2 = 1 #