G12 Adv S1 Unit 2-2024-2025 Worksheet PDF

Lesson 2-3 Computation of Limits Rules for calculating limits: 1) For any constant 𝑐 and any real number 𝑎 : lim 𝑐 = 𝑐 𝑥→𝑎 (the limit of constant equals the same constant) 2) For any real num...

Lesson 2-3 Computation of Limits Rules for calculating limits: 1) For any constant 𝑐 and any real number 𝑎 : lim 𝑐 = 𝑐 𝑥→𝑎 (the limit of constant equals the same constant) 2) For any real number 𝑎: lim 𝑥 = 𝑎 𝑥→𝑎 3) Suppose that lim 𝑓(𝑥) and lim 𝑔(𝑥) both are exist and let 𝐿, 𝑀, 𝑐 are real numbers 𝑥→𝑎 𝑥→𝑎 and lim 𝑓(𝑥) = 𝐿 and lim 𝑔(𝑥) = 𝑀 then the following apply: 𝑥→𝑎 𝑥→𝑎 a) lim 𝑐𝑓(𝑥) = 𝑐𝑎 Multiplying 𝑥→𝑎 with constant b) lim (𝑓(𝑥 ) ± 𝑔(𝑥 )) = lim 𝑓(𝑥) ± lim 𝑔(𝑥 ) = 𝐿 ± 𝑀 Addition and 𝑥→𝑎 𝑥→𝑎 𝑥→𝑎 subtraction c) lim (𝑓(𝑥 ) × 𝑔(𝑥 )) = lim 𝑓(𝑥) × lim 𝑔(𝑥 ) = 𝐿 × 𝑀 Multiplying 𝑥→𝑎 𝑥→𝑎 𝑥→𝑎 𝑓(𝑥) lim 𝑓(𝑥)± 𝐿 Dividing d) lim ( )= 𝑥→𝑎 = , 𝑖𝑓 lim 𝑔(𝑥 ) ≠ 0 𝑥→𝑎 𝑔(𝑥) lim 𝑔(𝑥) 𝑥→𝑎 𝑀 𝑥→𝑎 e) lim ( 𝑔(𝑥 ))𝑛 = (lim 𝑔(𝑥))𝑛 = 𝑀𝑛 Power of 𝑛 𝑥→𝑎 𝑥→𝑎 Radical f) lim √f(x) = √lim f(x) = √L ; 𝐿 ≥ 0 𝑓𝑜𝑟 𝑓(𝑥) ≥ 0 x→a x→a functions 𝑛 nth root g) lim 𝑛√𝑓(𝑥) = 𝑛√ lim 𝑓(𝑥) = √𝐿 ; 𝐿 ≥ 0 𝑤𝑒𝑟𝑒 𝑛 𝑒𝑣𝑒𝑛 x→a 𝑥→𝑎 Note: to find the limits, direct substitute in the given functions 14 | P a g e Use both graphs to find the limits if they exist: 25 g(x) f(x) 26 𝐥𝐢𝐦 𝒈(𝒙) = 27 𝐥𝐢𝐦 𝟐𝒇(𝒙) = 𝒙→𝟎 𝒙→𝟏 28 𝐥𝐢𝐦(𝒇(𝒙) + 𝟑𝒈(𝒙) − 𝟏) = 29 𝐥𝐢𝐦 ( 𝒈(𝒙) + 𝟑𝒙) = 𝒙→𝟐 𝒙→𝟏 30 𝐥𝐢𝐦 𝒙 𝒈(𝒙) = 31 𝒇(𝒙)+𝟏 𝒙→−𝟐 𝐥𝐢𝐦 ( )= 𝒙→𝟑 𝒈(𝒙) 15 | P a g e 32 a) 𝐥𝐢𝐦+ 𝒑(𝒙) = 𝒙→−𝟏 b) 𝐥𝐢𝐦 𝑵(𝒙) = 𝒓𝒆𝒂𝒔𝒐𝒏: … … …. 𝒙→−𝟏 c) 𝐥𝐢𝐦 𝑵(𝒙) = 𝒙→𝟏 d) 𝐥𝐢𝐦 𝒑(𝒙) = 𝒙→𝟏 e) 𝐥𝐢𝐦(𝒑(𝒙) + 𝑵(𝒙)) = 𝒙→𝟑 f) If 𝐥𝐢𝐦 𝒑(𝒙) = 𝟑 then the value of 𝒂 = 𝒙→𝒂 g) If 𝐥𝐢𝐦 𝑵(𝒙) = 𝟓 then the value of 𝒄 = 𝒙→𝒄 16 | P a g e Use 𝐥𝐢𝐦 𝒇(𝒙) = 𝟐 , 𝐥𝐢𝐦 𝒈(𝒙) = −𝟑 , 𝐥𝐢𝐦 𝒉(𝒙) = 𝟎 to determine the indicated limits, if 𝒙→𝒂 𝒙→𝒂 𝒙→𝒂 possible: 33 𝐥𝐢𝐦(𝟐𝒇(𝒙) − 𝟑𝒈(𝒙)) = 34 𝐥𝐢𝐦(𝟑𝒇(𝒙)𝒈(𝒙)) = 𝒙→𝒂 𝒙→𝒂 35 𝟐 36 𝟐𝒇(𝒙)𝒉(𝒙) 𝐥𝐢𝐦 ( (𝒇(𝒙)) )= 𝐥𝐢𝐦 ( )= 𝒈(𝒙) 𝒙→𝒂 𝒇(𝒙)+𝒉(𝒙) 𝒙→𝒂 17 | P a g e  Factorizing revision: Type: Example: Exercise: Common 𝟐𝟓𝒙 − 𝟏𝟎 = 𝟓(𝟓𝒙 − 𝟐) 𝟐𝒙𝟐 + 𝟔𝒙 = factored Difference 𝟐 𝟐 𝟒𝒙𝟐 − 𝟗 = of two 𝒙 − 𝒂 = (𝒙 − 𝒂)(𝒙 + 𝒂) squares 𝒂𝟐 − 𝒙𝟐 = (𝒂 − 𝒙)(𝒂 + 𝒙) 𝟐𝟓 − 𝟒𝒙𝟐 = Difference 𝒙𝟑 − 𝒂𝟑 = (𝒙 − 𝒂)(𝒙𝟐 + 𝒂𝒙 + 𝒂𝟐 ) 𝒙𝟑 − 𝟖 = of two cubes 𝒙𝟑 + 𝒂𝟑 = (𝒙 + 𝒂)(𝒙𝟐 − 𝒂𝒙 + 𝒂𝟐 ) 𝒙𝟑 + 𝟐𝟕= Polynomials 𝒙𝟐 + 𝟒𝒙 + 𝟑 = (𝒙 + 𝟑)(𝒙 + 𝟏) 𝒙𝟐 + 𝟓𝒙 − 𝟔 = 𝒙𝟐 − 𝟒𝒙 + 𝟑 = (𝒙 − 𝟑)(𝒙 − 𝟏) 𝟓𝒙𝟐 − 𝟕𝒙 + 𝟐 = 𝟐 𝒙 + 𝟒𝒙 − 𝟓 = (𝒙 + 𝟓)(𝒙 − 𝟏) 𝒙𝟐 − 𝟒𝒙 − 𝟓 = (𝒙 − 𝟓)(𝒙 + 𝟏) 𝒙𝟐 − 𝟑𝒙 − 𝟏𝟎 = 𝟑𝒙𝟐 + 𝟒𝒙 + 𝟏 = (𝟑𝒙 + 𝟏)(𝒙 + 𝟏) 𝟐 𝟐𝒙𝟐 + 𝟏𝟎𝒙 + 𝟏𝟐 = 𝟑𝒙 + 𝟑𝒙 − 𝟔 = 𝟑(𝒙 + 𝟏)(𝒙 − 𝟐) Groups 𝒙𝟑 − 𝟑𝒙𝟐 + 𝟒𝒙 − 𝟏𝟐 = 𝒙𝟑 + 𝟐𝒙𝟐 + 𝟐𝒙 + 𝟒 = (𝒙𝟑 + 𝟐𝒙𝟐 ) + (𝟐𝒙 + 𝟒) = 𝒙𝟐 (𝒙 + 𝟐) + 𝟐(𝒙 + 𝟐) = (𝒙 + 𝟐)(𝒙𝟐 + 𝟐) 18 | P a g e  First type of limits is polynomial functions limits (direct substitutions). Evaluate the indicated limits, if it exists: 37 𝐥𝐢𝐦 𝟑 = 38 𝒙→𝟐 𝐥𝐢𝐦√𝟒 = 𝒙→𝟒 39 𝐥𝐢𝐦 𝝅 = 40 𝐥𝐢𝐦 −𝟓𝒙 = 𝒙→−𝟐 𝒙→𝟒 41 𝐥𝐢𝐦(𝒙𝟐 − 𝟑𝒙 + 𝟏) = 42 𝐥𝐢𝐦 (−𝟓𝒙𝟑 − 𝟐𝒙𝟐 + 𝟏) = 𝒙→𝟎 𝒙→−𝟏 43 𝟑 44 𝟒 𝐥𝐢𝐦 √(𝟐𝒙 + 𝟏) = 𝐥𝐢𝐦 √(𝟐𝒙𝟐 + 𝒙 + 𝟏) = 𝒙→𝟐 𝒙→𝟑 45 𝐥𝐢𝐦(𝟑𝒙 − 𝟏)(𝒙 − 𝟓) = 46 𝐥𝐢𝐦(𝟐𝒙 + 𝟏)𝟑 = 𝒙→𝟑 𝒙→𝟐 47 𝒙+𝟑 48 𝒙𝟐 −𝟑𝒙−𝟏 𝐥𝐢𝐦 = 𝐥𝐢𝐦 = 𝒙→𝟏 𝟐𝒙+𝟕 𝒙→−𝟏 𝒙𝟐 +𝟏 19 | P a g e  Second type of limits is fraction functions limits. 𝒙𝟐 −𝟏 𝟎 If 𝐥𝐢𝐦 = then it is indeterminant, to solve it you must simplify the function 𝒙→𝟏 𝒙−𝟏 𝟎 by factorizing or by multiplying with the conjugate. Evaluate the indicated limits, if it exists: 49 𝒙𝟐 −𝒙−𝟔 50 𝒙𝟐 +𝒙−𝟐 𝐥𝐢𝐦 = 𝐥𝐢𝐦 = 𝒙→𝟑 𝒙−𝟑 𝒙→𝟏 𝒙𝟐 −𝟑𝒙+𝟐 51 𝒙𝟐 −𝒙−𝟐 52 𝒙𝟑 −𝟏 𝐥𝐢𝐦 = 𝐥𝐢𝐦 = 𝒙→𝟐 𝒙𝟐 −𝟒 𝒙→𝟏 𝒙𝟐 +𝟐𝒙−𝟑 53 𝒙𝟐 −𝟐𝒙+𝟏 54 𝒙𝟐 −𝒙 𝐥𝐢𝐦 = 𝐥𝐢𝐦 = 𝒙→𝟏 𝒙𝟐 −𝟏 𝒙→𝟏 𝒙𝟐 −𝟒𝒙+𝟑 20 | P a g e  Third type of limits is radical root functions limits. Note: conjugate product of square root is (√𝐟(𝐱) − 𝒂) (√𝒇(𝒙) + 𝒂) = (√𝒇(𝒙))𝟐 − 𝒂𝟐 = 𝒇(𝒙) − 𝒂𝟐 Ex.) (√𝐱 − 𝟐) (√𝒙 + 𝟐) = 𝒙 − 𝟒 Evaluate the indicated limits, if it exists: 55 √𝒙−𝟑 56 𝒙−𝟒 𝐥𝐢𝐦 𝐥𝐢𝐦 𝒙→𝟗 𝒙𝟐 −𝟗𝒙 𝒙→𝟒 √𝒙−𝟐 57 √𝒙+𝟒−𝟐 58 𝟐𝒙 𝐥𝐢𝐦 𝐥𝐢𝐦 𝒙→𝟎 𝒙 𝒙→𝟎 𝟑−√𝒙+𝟗 21 | P a g e 59 𝒙−𝒂−𝟑 60 𝒙𝟐 −𝟑𝒙 If 𝐥𝐢𝐦 √ 𝒙−𝟏 = 𝒆𝒙𝒊𝒔𝒕 𝒕𝒉𝒆𝒏 𝒇𝒊𝒏𝒅 𝒂 𝐥𝐢𝐦 𝒙→𝟏 𝒙→𝟑 𝟐−√𝒙+𝟏 61 𝒙−𝟏 62 𝒙𝟐 −𝟒 𝐥𝐢𝐦 𝐥𝐢𝐦 𝒙→𝟏 √𝒙−𝟏 𝒙→𝟐 𝟐−√𝟐𝒙 22 | P a g e 63 √𝒙+𝟐−√𝟐 64 √𝒙−𝟏 𝐥𝐢𝐦 𝐥𝐢𝐦 𝒙→𝟎 𝒙 𝒙→𝟏 𝟏−𝒙𝟐 Fourth type of limits is Piecewise-defined functions limits. 𝒇(𝒙) ,𝒙 ≤ 𝒂 𝐥𝐢𝐦− 𝒇(𝒙) , 𝒙 ≤ 𝒂 𝒊𝒇 𝒚 = { 𝒕𝒉𝒆𝒏 𝐥𝐢𝐦 𝒚 = { 𝒙→𝒂 𝒈(𝒙) ,𝒙 > 𝒂 𝒙→𝒂 𝐥𝐢𝐦+ 𝒈(𝒙) , 𝒙 > 𝒂 𝒙→𝒂 𝑖𝑓 lim− 𝑓(𝑥) = lim+ 𝑔(𝑥) 𝑡ℎ𝑒𝑛 𝑡ℎ𝑒 𝒍𝒊𝒎𝒊𝒕 𝒆𝒙𝒊𝒔𝒕𝒔 , 𝑜𝑡ℎ𝑒𝑟𝑤𝑖𝑠𝑒 𝒍𝒊𝒎𝒊𝒕 𝒅𝒐𝒆𝒔 𝒏𝒐𝒕 𝒆𝒙𝒊𝒔𝒕. 𝑥→𝑎 𝑥→𝑎 Evaluate the limit where “f(x)” defined at: 65 𝟑𝐱 − 𝟏; 𝐱 < 𝟐 𝒇(𝒙) = { 𝟔 ; 𝐱 = 𝟐 find the following limits 𝐱 𝟐 + 𝟏; 𝐱 > 𝟐 a) 𝐥𝐢𝐦 𝒇(𝒙) = b) 𝐥𝐢𝐦 𝒇(𝒙) = 𝒙→𝟑 𝒙→𝟏 c) 𝐥𝐢𝐦− 𝒇(𝒙) = d) 𝐥𝐢𝐦+ 𝒇(𝒙) = 𝒙→𝟐 𝒙→𝟐 e) 𝐥𝐢𝐦 𝒇(𝒙) = f) 𝒇(𝟐) = 𝒙→𝟐 23 | P a g e 66 𝟐𝒙 , 𝒙 > 𝟐 𝒇(𝒙) = { 𝟐 find 𝐥𝐢𝐦 𝒇(𝒙) 𝒙 ,𝒙 ≤ 𝟐 𝒙→𝟐 67 𝒙𝟐 + 𝟏 , 𝒙 < −𝟏 𝒇(𝒙) = { find 𝐥𝐢𝐦 𝒇(𝒙) 𝟑𝒙 + 𝟏 , 𝒙 > −𝟏 𝒙→−𝟏 68 𝟐𝒙 + 𝟏 , 𝒙 < −𝟏 { 𝟑 , −𝟏 < 𝒙 < 𝟏 find a) 𝐥𝐢𝐦 𝒇(𝒙) b) 𝐥𝐢𝐦 𝒇(𝒙) 𝒙→−𝟏 𝒙→𝟏 𝟐𝒙 + 𝟏 ,𝒙 > 𝟏 24 | P a g e  Fifth type of limits is Absolute Value Functions limits Absolute value function is a continuous function and the limit of this function exists on all real numbers when it’s a single function, but when its merged with other function like rational functions. then it must be rewritten as a Piecewise- defined function and then check the limit on both sides. left and right sides. 𝒇(𝒙) 𝒇(𝒙) > 𝟎 𝒚 = |𝒇(𝒙)| = { −𝒇(𝒙) 𝒇(𝒙) ≤ 𝟎 Evaluate the indicated limit, if it exists: 69 𝐥𝐢𝐦|𝒙| = 70 𝐥𝐢𝐦|𝟑𝒙 − 𝟕| = 𝒙→𝟎 𝒙→𝟏 71 |𝒙+𝟏|−𝟐 72 |𝒙−𝟐|−𝟏 𝐥𝐢𝐦 = 𝐥𝐢𝐦 = 𝒙→𝟏 𝒙𝟐 −𝒙 𝒙→𝟏 𝒙𝟐 −𝒙 25 | P a g e 73 |𝒙| 74 |𝒙| 𝐥𝐢𝐦− = 𝐥𝐢𝐦 = 𝒙→𝟎 𝒙 𝒙→𝟎 𝒙 75 𝒙𝟐 −𝒃𝟐 𝐥𝐢𝐦 = 𝐥𝐢𝐦 (𝒙|𝒙| + 𝟔) find the value of 𝒃 𝒙→𝒃 𝒙−𝒃 𝒙→−𝟐 26 | P a g e  Sixth type of limits is the limit of the greatest integer number. Definition [𝒙] = 𝒂 , 𝒂 ≤ 𝒙 < 𝒂 + 𝟏 𝒚 = [ 𝒙] Ex.) [𝟖. 𝟏𝟐] = 𝟖 , [−𝟐. 𝟒] = −𝟑 [𝟑. 𝟗] = 𝟑 , [−𝟑. 𝟗𝟗] = −𝟒 Evaluate the indicated limits, if it exist: 76 𝐥𝐢𝐦 [𝒙] = 77 𝐥𝐢𝐦 [𝒙] = 𝒙→𝟑− 𝒙→𝟑 78 𝐥𝐢𝐦−([𝒙] + 𝟐𝒙) = 79 𝐥𝐢𝐦( [𝒙] + 𝟒𝒙 + 𝟏) = 𝒙→𝟑 𝟏 𝒙→𝟐 80 𝐥𝐢𝐦 [𝟐𝒙] = 81 𝒙𝟐 −[𝒙] 𝒙→𝟏.𝟓 𝐥𝐢𝐦+ |𝒙−𝟐|−𝟏 𝒙→𝟏 27 | P a g e  Seventh type of limits is limit of trigonometry and inverse trigonometry functions. 𝒔𝒊𝒏 𝒙 𝒙 𝒕𝒂𝒏 𝒙 𝒙 𝐥𝐢𝐦 = 𝟏 , 𝐥𝐢𝐦 = 𝟏 , 𝐥𝐢𝐦 = 𝟏 , 𝐥𝐢𝐦 =𝟏, 𝒙→𝟎 𝒙 𝒙→𝟎 𝒔𝒊𝒏 𝒙 𝒙→𝟎 𝒙 𝒙→𝟎 𝒕𝒂𝒏 𝒙 𝒔𝒊𝒏 𝒂𝒙 𝒂 𝒃𝒙 𝒃 𝒕𝒂𝒏 𝒂 𝒙 𝒂 𝒃𝒙 𝒃 𝐥𝐢𝐦 = , 𝐥𝐢𝐦 = , 𝐥𝐢𝐦 = , 𝐥𝐢𝐦 = , 𝒙→𝟎 𝒃𝒙 𝒃 𝒙→𝟎 𝒔𝒊𝒏 𝒂𝒙 𝒂 𝒙→𝟎 𝒃𝒙 𝒃 𝒙→𝟎 𝒕𝒂𝒏 𝒂 𝒙 𝒂 Evaluate the indicated limits, if it exists: 82 𝒔𝒊𝒏 𝒙 83 𝒔𝒊𝒏 𝒙 𝐥𝐢𝐦 = 𝐥𝐢𝐦 ( )= 𝒙→𝟎 𝟑𝒙 𝒙→𝟎 𝒕𝒂𝒏 𝒙 84 𝒕𝒂𝒏𝟐 𝒙 85 𝒔𝒊𝒏 𝟓𝒙 𝐥𝐢𝐦 = 𝐥𝐢𝐦 = 𝒙→𝟎 𝒙 𝒙→𝟎 𝟐𝒙 86 𝒙 87 𝒔𝒊𝒏𝟐 𝟑𝒙 𝐥𝐢𝐦 = 𝐥𝐢𝐦 = 𝒙→𝟎 𝒔𝒊𝒏(𝟒𝒙) 𝒙→𝟎 𝟐𝒙𝟐 88 𝒕𝒂𝒏𝟑𝒙+𝟐𝒙 𝒔𝒊𝒏𝒙 89 𝐥𝐢𝐦 𝒕𝒂𝒏𝟐𝒙 𝒄𝒔𝒄𝝅𝒙 = 𝐥𝐢𝐦 = 𝒙→𝟎 𝒙→𝟎 𝟓𝒙 28 | P a g e 90 𝒕𝒂𝒏𝟓𝒙 𝒄𝒐𝒔𝟑𝒙 91 𝒔𝒊𝒏𝟐𝒙 𝐥𝐢𝐦 = 𝐥𝐢𝐦 = 𝒙→𝟎 𝟐𝒙 𝒙→𝟎 𝒙𝟐 −𝟐𝒙 92 𝒔𝒊𝒏𝟐 𝟑𝒙− 𝒕𝒂𝒏𝟐𝒙 93 𝒔𝒊𝒏𝟐 𝟑𝒙 𝐥𝐢𝐦 = 𝐥𝐢𝐦− = 𝒙→𝟎 𝒙𝒄𝒐𝒔𝒙+𝟒 𝒕𝒂𝒏𝒙 𝒙→𝟎 𝟑𝒙|𝒙| 29 | P a g e 94 𝒔𝒊𝒏𝟐𝒙 95 𝟏−𝒄𝒐𝒔𝟐𝒙 𝐥𝐢𝐦− = 𝐥𝐢𝐦 = 𝒙→𝟎 𝟑𝒙[𝒙] 𝒙→𝟎 𝟓𝒙𝟐 96 𝒔𝒊𝒏𝒙 97 𝐥𝐢𝐦 𝟓𝒙( 𝒄𝒔𝒄𝟐𝒙 + 𝒄𝒐𝒕𝒙) = 𝐥𝐢𝐦 = 𝒙→𝟎 𝒙→𝟎 𝟏−√𝒙+𝟏 30 | P a g e  98 𝐥𝐢𝐦 ( 𝒙𝟐 𝐜𝐬𝐜 𝟐 𝒙) = 99 𝒔𝒊𝒏 |𝒙| 𝒙→𝟎 𝐥𝐢𝐦 ( )= 𝒙→𝟎 𝒙 100 𝒕𝒂𝒏 𝒙.𝒄𝒐𝒔 𝒙 101 𝒔𝒊𝒏 (𝒙𝟐 −𝟒) 𝐥𝐢𝐦 ( )= 𝐥𝐢𝐦 ( )= 𝒙→𝟎 𝟓𝒙 𝒙→𝟐 𝒙𝟐 −𝟒 31 | P a g e 102 𝟏−𝒄𝒐𝒔𝒙 103 𝐥𝐢𝐦 𝐥𝐢𝐦 (𝟐𝒙(√𝟏 + 𝒄𝒐𝒕𝟐 𝒙) 𝒙→𝟎 𝒙𝟐 𝒙→𝟎− 104 𝒙+𝟏 105 𝐥𝐢𝐦 𝒄𝒐𝒔−𝟏 (𝒙𝟐 ) = 𝐥𝐢𝐦 𝐬𝐢𝐧−𝟏 ( )= 𝒙→𝟎 𝒙→𝟎 𝟐 32 | P a g e 106 𝟐 𝟐 𝐥𝐢𝐦 ( − |𝒙| ) 𝒙→𝟎 𝒙 107 𝟏 𝟐 𝐥𝐢𝐦 ( − ) 𝒙→𝟏 𝒙−𝟏 𝒙𝟐 −𝟏 108 𝟏 𝟏 𝒍𝒊𝒎 ( 𝒙 − 𝒙𝟐 +𝒙 ) 𝒙→𝟎 33 | P a g e 109 𝒔𝒊𝒏𝒙 110 𝒙𝒆𝟐𝒙+𝟏 𝐥𝐢𝐦+ 𝐥𝐢𝐦 𝒙𝟐 +𝒙 𝒙→𝟎 √𝒙 𝒙→𝟎 If 𝒑(𝒙) = 𝒙𝟐 − 𝟏 answer Q 117,118: 111 𝐥𝐢𝐦(𝒑 (𝒑(𝒑(𝒑(𝒙)))) = 𝒙→𝟎 112 𝐥𝐢𝐦(𝒑(𝟑 + 𝟐𝒑(𝒙 − 𝒑(𝒙))) = 𝒙→𝟎 113 (𝟐+𝒉)𝟐 −𝟒 𝐥𝐢𝐦 𝒉→𝟎 𝒉 34 | P a g e 114 𝒂 + 𝟎, 𝟏𝟐𝒙 𝒙 ≤ 𝟐𝟎, 𝟎𝟎𝟎 𝑻(𝒙) = { 𝒃 + 𝟎. 𝟏𝟔(𝒙 − 𝟐𝟎, 𝟎𝟎𝟎𝟎) 𝒙 > 𝟐𝟎, 𝟎𝟎𝟎 find the value of 𝒂, 𝒃 therefore. 𝐥𝐢𝐦 𝑻(𝒙) 𝒆𝒙𝒊𝒔𝒕 , 𝒂𝒏𝒅 𝐥𝐢𝐦+ 𝑻(𝒙) = 𝟎 𝒙→𝟐𝟎𝟎𝟎𝟎 𝒙→𝟎 115 |𝒙+𝟏|−𝟐 𝟓𝒙 𝐥𝐢𝐦+ = 𝐥𝐢𝐦 𝒔𝒊𝒏 𝒌𝒙 find the value of 𝒌 𝒙→𝟏 𝒙𝟐 −𝒙 𝒙→𝟎 35 | P a g e  Eighth: squeeze Theorem Use the squeeze theorem to verify the values of limits: 116 𝟏 𝐥𝐢𝐦 𝒙𝟐 𝒔𝒊𝒏 𝒙→𝟎 𝒙 117 𝟏 𝐥𝐢𝐦(𝟓 − 𝒙𝟐 𝒄𝒐𝒔 ) 𝒙→𝟎 𝒙 36 | P a g e 118 If (𝒙𝟐 − 𝟏)(𝒙𝟐 + 𝟏) ≤ (𝒙 − 𝟏)𝒇(𝒙) ≤ 𝒙𝟐 + 𝟐𝒙 − 𝟑 𝒙 ≠ 𝟏 𝒂𝒏𝒅 𝒙 ∈ [−𝟑, 𝟑] find 𝐥𝐢𝐦 𝒇(𝒙) 𝒙→𝟏 119 if 𝒔𝒊𝒏 𝒙 + 𝒙 ≤ 𝒇(𝒙) ≤ 𝒙𝟐 + 𝟐𝒙 𝒇(𝒙) 𝒙 ≠ 𝟎 𝒂𝒏𝒅 𝒙 ∈ [−𝝅, 𝝅] find 𝐥𝐢𝐦 𝒙→𝟎 𝒙 37 | P a g e  Use of limits in computing velocity. we see that for an object moving in straight line , whose position at time 𝒕 is given by the function 𝒇(𝒕) , the instantaneous velocity of that object at time 𝒕 𝒇(𝒕+𝒉)−𝒇(𝒕) the average velocity over some period is given by the limit 𝐥𝐢𝐦 𝒉→𝟎 𝒉 Use the given position function 𝒇(𝒕) to find the velocity at time 𝒕 = 𝒂: 120 𝒇(𝒕) = 𝒕𝟐 + 𝟐 𝒂=𝟐 121 𝒇(𝒕) = 𝒕𝟑 + 𝟐 𝒂=𝟎 38 | P a g e  TRY BY Your Self (2) 1 Evaluate 𝐥𝐢𝐦(𝒙𝟐 − 𝟑𝒙 + 𝟏) 𝒙→𝟎 A 2 B 0 C -1 D Does not exist 2 𝒙𝟐 −𝟏 Evaluate 𝐥𝐢𝐦( ) 𝒙→𝟎 𝒙+𝟏 A 2 B 0 C -1 D Does not exist 3 Evaluate 𝐥𝐢𝐦( 𝒙+𝟒 ) 𝒙→𝟐 𝒙𝟐 +𝟐𝒙+𝟏 A 2 B 0 C -1 D Does not exist 4 Evaluate 𝐥𝐢𝐦 (√𝒙𝟐 + 𝟒) 𝒙→−𝟐 A 2 B 0 C -1 D Does not exist 5 𝒙𝟐 −𝟏 Evaluate 𝐥𝐢𝐦 ( ) 𝒙→−𝟏 𝒙+𝟏 A 2 B 0 C -1 D Does not exist 39 | P a g e 6 √𝟐𝒙+𝟏+𝟑 Evaluate 𝐥𝐢𝐦( ) 𝒙→𝟒 𝒙−𝟐 A 2 B 0 C -1 D Does not exist 7 𝒙𝟑 −𝟏 Evaluate 𝐥𝐢𝐦( ) 𝒙→𝟏 𝒙𝟐 +𝟐𝒙−𝟑 A 2 B 0 C -1 D Does not exist 8 Evaluate 𝐥𝐢𝐦(𝒔𝒊𝒏−𝟏 (𝒙𝟐 ) 𝒙→𝟎 A 2 B 0 C -1 D Does not exist 9 Evaluate 𝐥𝐢𝐦( 𝒕𝒂𝒏𝟑𝒙 ) 𝒙→𝟎 𝟐𝒔𝒊𝒏𝟓𝒙 A 2 B 0 C -1 D Does not exist 10 Evaluate 𝐥𝐢𝐦(𝒙𝟐 𝒄𝒔𝒄𝟐 𝒙) 𝒙→𝟎 A 2 B 0 C -1 D Does not exist 40 | P a g e 11 𝒙𝒆𝟑𝒙+𝟏 Evaluate 𝐥𝐢𝐦( ) 𝒙→𝟎 𝒙𝟐 +𝒙 A 2 B 0 C -1 D Does not exist 12 √𝒙+𝟒−𝟐 Evaluate 𝐥𝐢𝐦( ) 𝒙→𝟎 𝒙 A 2 B 0 C -1 D Does not exist 13 𝒙𝟐 −𝟏 Evaluate 𝐥𝐢𝐦( ) 𝒙→𝟎 𝒙+𝟏 A 2 B 0 C -1 D Does not exist 14 Evaluate 𝐥𝐢𝐦( 𝒙−𝟏 ) 𝒙→𝟏 √𝒙−𝟏 A 2 B 0 C -1 D Does not exist 15 𝒙𝟑 −𝟐𝟕 Evaluate 𝐥𝐢𝐦( ) 𝒙→𝟑 𝒙𝟐 −𝟐𝒙−𝟑 A 2 B 0 C -1 D Does not exist 41 | P a g e 16 Evaluate 𝟐 𝐥𝐢𝐦( − |𝒙|) 𝟐 𝒙→𝟎 𝒙 A 2 B 0 C -1 D Does not exist 17 𝟏−𝒆𝟐𝒙 Evaluate 𝐥𝐢𝐦( ) 𝒙→𝟎 𝟏−𝒆𝒙 A 2 B 0 C -1 D Does not exist 18 𝒔𝒊𝒏|𝒙| Evaluate 𝐥𝐢𝐦( ) 𝒙→𝟎 𝒙 A 2 B 0 C -1 D Does not exist 19 𝒙𝟐 Evaluate 𝐥𝐢𝐦( ) 𝒙→𝟎 𝒙𝟑 +𝒙 A 2 B 0 C -1 D Does not exist 20 Evaluate 𝐥𝐢𝐦( 𝒄𝒐𝒔𝒙−𝟏 ) 𝒙→𝟎 𝒙 A 2 B 0 C -1 D Does not exist 42 | P a g e  Lesson 2-4 Continuity and its Consequence The function will be continuous on point 𝒙 = 𝒄 if and only if the following conditions applied: 1) 𝒇(𝒄) = 𝑳 (𝒅𝒆𝒇𝒊𝒏𝒆𝒅) 2) 𝐥𝐢𝐦 𝒇(𝒙) = 𝑳 (𝒆𝒙𝒊𝒔𝒕) 𝒙→𝒄 3) 𝐥𝐢𝐦 𝒇(𝒙) = 𝒇(𝒄) = 𝑳 𝒙→𝒄 Discontinuous cases: - 1.) Removable discontinuous: lim 𝑓(𝑥 ) = 𝑒𝑥𝑖𝑠𝑡 𝑥→𝑎 lim 𝑓 (𝑥 ) = 𝑒𝑥𝑖𝑠𝑡 𝑓(𝑎) = 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑥→𝑎 𝑓(𝑎) = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑 lim 𝑓(𝑥 ) ≠ 𝑓(𝑎) 𝑥→𝑎 The graph of the function has a hole at The graph of the function has a hole at 𝑥 = 𝑥=𝑎 43 | P a g e  2.) Nonremovable discontinuous: lim 𝑓(𝑥 ) = 𝑑𝑜𝑒 𝑛𝑜𝑡 𝑒𝑥𝑖𝑠𝑡 lim 𝑓(𝑥 ) = 𝑑𝑜𝑒 𝑛𝑜𝑡 𝑒𝑥𝑖𝑠𝑡 𝑥→𝑎 𝑥→𝑎 𝑓(𝑎) = 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑓(𝑎) = 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 The graph of the function has a Jump The graph of the function has infinity at 𝑥 = 𝑎 (blows up) at 𝑥 = 𝑎 𝟏 𝒍𝒊𝒎 𝒄𝒐𝒔 (𝒙) = 𝒅𝒐𝒆𝒔 𝒏𝒐𝒕 𝒆𝒙𝒊𝒔𝒕 𝒙→𝟎 𝟏 the endless oscillation of 𝒄𝒐𝒔( ) 𝒙 44 | P a g e Use the given graphs to identify all intervals on which the function is continuous and the points the function is discontinuous and type of discontinuity: 122 123 45 | P a g e 124 Complete the table below: Type of discontinuity Reason: Function: at 𝒙 = 𝟎: 𝟏 𝒇(𝒙) = 𝒔𝒊𝒏 𝒙𝟐 𝟏 𝒇(𝒙) = 𝒙 𝟐−𝒙 𝒇(𝒙) = 𝒙 𝟐 𝒇 (𝒙 ) = { 𝒙 − 𝟓 𝒙>𝟎 𝒙 + 𝒄𝒐𝒔 𝒙 𝒙 ≤ 𝟎 𝒇 (𝒙 ) = { √ 𝒙 + 𝟒 𝟐 𝒙≠𝟎 𝟒 𝒙=𝟎 46 | P a g e  Type of Function: Point of discontinuity: discontinuity: 𝒙𝟐 − 𝟐𝒙 − 𝟑 𝒇(𝒙) = 𝒙−𝟑 𝒔𝒊𝒏𝟓𝒙 𝒇(𝒙) = 𝒙 𝟐 𝒇(𝒙) = 𝒙−𝟑 |𝒙| 𝒇(𝒙) = 𝒙 𝒙𝟐 − 𝟓𝒙 − 𝟔 𝒇(𝒙) = 𝒙−𝟔 𝒙−𝟐 𝒇(𝒙) = 𝒙𝟐 − 𝟒 𝒙−𝟓 𝒙>𝟐 𝒇(𝒙) = { 𝒙𝟐 + 𝟑 𝒙≤𝟐 Functions continuous everywhere: 1) Polynomials functions. 2) Trigonometry functions (𝒔𝒊𝒏 𝒙 , 𝒄𝒐𝒔 𝒙 , 𝒕𝒂𝒏−𝟏 𝒙). 3) Exponential functions (𝒚 = 𝒆𝒙 ). 4) Square roots (𝒚 = √𝒇(𝒙) , 𝒇(𝒙) ≥ 𝟎). 𝒏 5) √𝒇(𝒙) 𝒄𝒐𝒏𝒕𝒊𝒏𝒖𝒐𝒖𝒔 𝒐𝒏 𝒂𝒍𝒍 𝒙 𝒗𝒂𝒍𝒖𝒆𝒔 𝒘𝒉𝒆𝒏 "𝒏" 𝒊𝒔 𝒐𝒅𝒅 , 𝒇(𝒙) ≥ 𝟎 𝒘𝒉𝒆𝒏 is 𝒆𝒗𝒆𝒏. 6) Logarithm function (𝒚 = 𝒍𝒏(𝒇 (𝒙)) , 𝒇(𝒙) ≥ 𝟎). 7) Absolute value. 8) 𝒔𝒊𝒏−𝟏 𝒙 , 𝒄𝒐𝒔−𝟏 𝒙 𝒄𝒐𝒏𝒕𝒊𝒏𝒖𝒐𝒖𝒔 𝒐𝒏 − 𝟏 < 𝒙 < 𝟏. 47 | P a g e  9) 𝐥𝐢𝐦 𝒈(𝒙) = 𝑳 𝒂𝒏𝒅 𝒇 𝒄𝒐𝒏𝒕𝒊𝒏𝒖𝒐𝒖𝒔 𝒐𝒏 𝑳 𝒕𝒉𝒆𝒏: 𝒙→𝒂 𝐥𝐢𝐦 𝒇(𝒈(𝒙)) = 𝒇(𝐥𝐢𝐦 𝒈(𝒙)) = 𝒇(𝑳) 𝒙→𝒂 𝒙→𝒂 First: continuity for rational functions. Note: To remove a hole from a graph, the function must be extended to be continuous so rewrite the function as: 𝑓(𝑥 ) 𝑥 ≠ 𝑐 𝑓(𝑥 ) = { 𝑤ℎ𝑒𝑟𝑒 lim 𝑓(𝑥 ) = 𝑎 𝑎 𝑥=𝑐 𝑥→𝑐 Determine where 𝒇 is continuous, if possible, extend 𝒇 as a new function that is continuous on a larger domain: 125 𝒙𝟐 +𝒙−𝟐 126 𝒙𝟐 −𝒙−𝟔 𝒇(𝒙) = 𝒇(𝒙) = 𝒙+𝟐 𝒙−𝟑 48 | P a g e 127 𝒙−𝟏 128 𝟒𝒙 𝒇(𝒙) = 𝒇(𝒙) = 𝒙𝟐 −𝟏 𝒙𝟐 +𝒙−𝟐 129 𝟒𝒙 130 𝒙𝟐 −𝟐𝒙+𝟏 𝒇(𝒙) = 𝒇(𝒙) = 𝒙𝟐 +𝟒 𝒙𝟐 −𝟏 49 | P a g e 131 𝒔𝒊𝒏 𝟐𝒙−𝒕𝒂𝒏 𝒙 132 √𝒙+𝟏−𝟑 𝒇(𝒙) = 𝒇(𝒙) = 𝒙 𝒙−𝟖 133 𝒙+𝟑 134 √𝒙+𝟒−𝟐 𝒇(𝒙) = 𝟏 𝟏 𝒇(𝒙) = + 𝒔𝒊𝒏 𝒙 𝒙 𝟑 50 | P a g e 135 𝟏−𝒙𝟒 136 𝒆𝟐𝒙 −𝟏 𝒇(𝒙) = 𝒇(𝒙) = 𝒙𝟐 −𝟏 𝒆𝒙 −𝟏 137 |𝒙−𝟐|−𝟏 138 |𝒙−𝟏|−𝟏 𝒇(𝒙) = 𝒇(𝒙) = 𝒙−𝟑 𝒙 51 | P a g e  Second: continuity for Piecewise-defined functions. Note: to determine the continuity of a Piecewise-defined function 𝒇(𝒙) = 𝒉(𝒙) ,𝒙 ≤ 𝒂 { , the following must be applied.: 𝒈(𝒙) , 𝒙 > 𝒂 1) 𝑓(𝑥) = 𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑎𝑡 𝑥 = 𝑎 2) lim− 𝑓(𝑥) = lim+ 𝑓(𝑥) 𝑥→𝑎 𝑥→𝑎 3) 𝑓(𝑎) = lim 𝑓(𝑥) 𝑥→𝑎 Determine where 𝒇 is continuous, at point 𝒙 = 𝒂: 139 𝟐𝒙 𝒙 < 𝟏 140 𝟑𝒙 − 𝟏 𝒙 ≤ −𝟏 𝒇(𝒙) = { 𝟐 𝒙 𝒙≥𝟏 𝒇(𝒙) = {𝒙𝟐 + 𝟓𝒙 − 𝟏 < 𝒙 < 𝟏 𝟑𝒙𝟑 𝒙≥𝟏 141 𝟐𝒙 𝒙≤𝟎 142 𝒔𝒊𝒏 𝒙 𝒙≠𝟎 𝒇(𝒙) = { 𝒔𝒊𝒏 𝒙 𝟎 < 𝒙 < 𝝅 𝒇(𝒙) = { 𝒙 𝒙−𝝅 𝒙≥𝝅 𝟏 𝒙=𝟎 52 | P a g e 143 𝟏 144 𝟏 𝒙𝟐 𝒔𝒊𝒏 𝒙≠𝟎 𝒙𝟎 177 Determine the values of 𝒂 & 𝒃 that make the given function continuous on its domain: 𝟐𝒔𝒊𝒏 𝒙 𝒙𝟎 63 | P a g e 178 Determine values of 𝒂 & 𝒃 that make the given function continuous on its domain: 𝒂𝒆𝒙 + 𝟏 𝒙𝟐 179 Determine values of 𝒂 & 𝒃 that make the given function continuous on its domain: 𝒂(𝒕𝒂𝒏−𝟏 𝒙 + 𝟐) 𝒙𝟑 64 | P a g e 180 Prove that if f is continuous on the interval [𝒂, 𝒃], 𝒇(𝒂) > 𝒂, 𝒇(𝒃) < 𝒃, then f has a fixed point ( a solution of 𝒇(𝒙) = 𝒙 ) in the interval (𝒂, 𝒃): 181 𝒙 𝟐 𝒙 ≠ 𝟎 and 𝒈(𝒙) = 𝟐𝒙 If 𝒇(𝒙) = { 𝟒 𝒙=𝟎 show that 𝐥𝐢𝐦 𝒇(𝒈(𝒙)) ≠ 𝒇(𝐥𝐢𝐦 𝒈(𝒙)) : 𝒙→𝟎 𝒙→𝟎 65 | P a g e  TRY BY Your Self (4) 1 Determine the interval(s) where 𝒇(𝒙) = √𝒙 + 𝟐 is continuous. A 2 B 0 C -1 D Does not exist 2 Determine the interval(s) where 𝒇(𝒙) = √𝒙𝟐 − 𝟗 is continuous. A 2 B 0 C -1 D Does not exist 3 Determine the interval(s) where 𝟑 𝒇(𝒙) = √𝒙𝟐 − 𝟒 is continuous. A 2 B 0 C -1 D Does not exist 4 𝟑 Determine the interval(s) where 𝒇(𝒙) = (𝒙 − 𝟏) is continuous. 𝟐 A 2 B 0 C -1 D Does not exist 5 Determine the interval(s) where 𝒇(𝒙) = 𝒔𝒊𝒏−𝟏 (𝒙) is continuous. A 2 B 0 C -1 D Does not exist 66 | P a g e 6 Determine the interval(s) where 𝒍𝒏(𝒔𝒊𝒏 𝒙) is continuous. A 2 B 0 C -1 D Does not exist 7 √𝑿+𝟐+𝒆𝒙 Determine the interval(s) where 𝒇(𝒙) = is continuous. 𝒙𝟐 −𝟒 A 2 B 0 C -1 D Does not exist 8 𝐥𝐧(𝐱 𝟐 −𝟏 Determine the interval(s) where 𝒇(𝒙) = is continuous. √𝒙𝟐 −𝟐𝒙 A 2 B 0 C -1 D Does not exist 9 Determine the interval(s) where 𝒇(𝒙) = √𝒙 + 𝟐 is continuous. A 2 B 0 C -1 D Does not exist 10 𝟗−𝒙𝟐 If 𝒇(𝒙) = is continuous. On (−∞, ∞) 𝒇𝒊𝒏𝒅 𝒎, where m is constant 𝒎𝒙+𝟐 number A 0 B -2 C 1 D 3 67 | P a g e  Lesson 2-5 Limits Involving Infinity, Asymptotes 𝟏 Let 𝒇(𝒙) = , the table shows that: 𝒙 𝒙 -0.001 -0.0001 -0.00001 0 0.00001 0.0001 0.001 𝒇(𝒙) -1000 -10000 -100000 100000 10000 1000 when "𝒙" is approaching to zero from right, 𝒇(𝒙) is approaching to unbounded positive number and when "𝒙" is approaching to zero from left 𝒇(𝒙) is approaching to unbounded negative number. 𝟏 𝟏 That means 𝐥𝐢𝐦+ → ∞ and 𝐥𝐢𝐦− → −∞ 𝒙→𝟎 𝒙 𝒙→𝟎 𝒙 𝟏 so 𝐥𝐢𝐦 = 𝒅𝒐𝒆𝒔 𝒏𝒐𝒕 𝒆𝒙𝒊𝒔𝒕 𝒙→𝟎 𝒙 𝟏 Let 𝒇(𝒙) = , complete the table and the indicated limits below: 𝒙𝟐 𝒙 -0.001 -0.0001 -0.00001 0 0.00001 0.0001 0.001 𝒇(𝒙) Find: 𝟏 1) 𝐥𝐢𝐦+ = 𝒙→𝟎 𝒙𝟐 𝟏 2) 𝐥𝐢𝐦− = 𝒙→𝟎 𝒙𝟐 𝟏 3) 𝐥𝐢𝐦 = 𝒙→𝟎 𝒙𝟐 68 | P a g e  𝒇(𝒙) Notes: when 𝐥𝐢𝐦 , 𝒈(𝒄) = 𝟎 , 𝒇(𝒄) ≠ 𝟎 there’s three prospects for the 𝒙→𝒄 𝒈(𝒙) answer: 𝒇(𝒙) Prospect 1: 𝐥𝐢𝐦 =∞ 𝒙→𝒄 𝒈(𝒙) 𝒇(𝒙) Prospect 2: 𝐥𝐢𝐦 = −∞ 𝒙→𝒄 𝒈(𝒙) 𝒇(𝒙) 𝒇(𝒙) 𝒇(𝒙) Prospect 3: 𝐥𝐢𝐦 𝒈(𝒙) = 𝒅𝒐𝒆𝒔 𝒏𝒐𝒕 𝒆𝒙𝒊𝒔𝒕 , 𝒃𝒆𝒄𝒂𝒖𝒔𝒆 𝐥𝐢𝐦− 𝒈(𝒙) = −∞ , 𝐥𝐢𝐦+ 𝒈(𝒙) = ∞ 𝒙→𝒄 𝒙→𝒄 𝒙→𝒄 Note: Determinant quantities: 𝒂 𝒂 1) = ±∞ , =𝟎 𝒂≠𝟎 𝟎 ±∞ 2) ∞ ± 𝒂 = ∞ , ∞+∞=∞ 3) 𝒂 × ∞ = ∞ 𝒊𝒇 𝒂 > 𝟎 , 𝒂 × ∞ = −∞ 𝒊𝒇 𝒂 < 𝟎 4) 𝒂𝟎 = 𝟏 , ∞∞ = ∞ 5) 𝒂∞ = ∞ , 𝒊𝒇 𝒂 > 𝟏 , 𝒂∞ = 𝟎 , 𝒊𝒇 𝟎 < 𝒂 < 𝟏 Indeterminant quantities: 𝟎 ±∞ , ,∞ − ∞ , 𝟎 × ∞ , 𝟎𝟎 , ∞𝟎 , 𝟏∞ 𝟎 ±∞ 69 | P a g e Determine: a) 𝐥𝐢𝐦+ 𝒇(𝒙) b) 𝐥𝐢𝐦− 𝒇(𝒙) c) 𝐥𝐢𝐦 𝒇(𝒙) (the answer will be 𝒙→𝒂 𝒙→𝒂 𝒙→𝒂 𝒏𝒖𝒎𝒃𝒆𝒓 , ∞ , −∞ 𝒐𝒓 𝒅𝒐𝒆𝒔 𝒏𝒐𝒕 𝒆𝒙𝒊𝒔𝒕 ): 182 𝒙 183 𝒙 𝒇(𝒙) = 𝒂𝒕 𝒂 = 𝟏 𝒇 (𝒙) = (𝒙+𝟏)𝟐 𝒂𝒕 𝒂 = −𝟏 𝒙−𝟏 𝒙 𝒙 a) 𝐥𝐢𝐦+ = a) 𝐥𝐢𝐦+ (𝒙+𝟏)𝟐 = 𝒙→𝟏 𝒙−𝟏 𝒙→−𝟏 𝒙 𝒙 b) 𝐥𝐢𝐦− = b) 𝐥𝐢𝐦− (𝒙+𝟏)𝟐 = 𝒙→𝟏 𝒙−𝟏 𝒙→−𝟏 𝒙 𝒙 c) 𝐥𝐢𝐦 = c) 𝐥𝐢𝐦 = 𝒙→𝟏 𝒙−𝟏 𝒙→−𝟏 (𝒙+𝟏)𝟐 184 𝟏−𝟐𝒙 185 𝟏−𝟐𝒙 𝒇(𝒙) = 𝒂𝒕 𝒂 = 𝟏 𝒇(𝒙) = 𝒂𝒕 𝒂 = −𝟏 𝒙𝟐 −𝟏 𝒙𝟐 −𝟏 𝟏−𝟐𝒙 𝟏−𝟐𝒙 d) 𝐥𝐢𝐦+ = d) 𝐥𝐢𝐦+ = 𝒙→𝟏 𝒙𝟐 −𝟏 𝒙→−𝟏 𝒙𝟐 −𝟏 𝟏−𝟐𝒙 𝟏−𝟐𝒙 e) 𝐥𝐢𝐦− = e) 𝐥𝐢𝐦− = 𝒙→𝟏 𝒙𝟐 −𝟏 𝒙→−𝟏 𝒙𝟐 −𝟏 𝟏−𝟐𝒙 𝟏−𝟐𝒙 f) 𝐥𝐢𝐦 = f) 𝐥𝐢𝐦 = 𝒙→𝟏 𝒙𝟐 −𝟏 𝒙→−𝟏 𝒙𝟐 −𝟏 70 | P a g e 186 𝒙−𝟒 187 𝟏−𝒙 𝒇(𝒙) = 𝒂=𝟐 𝒇(𝒙) = (𝒙+𝟏)𝟐 𝒂 = −𝟏 𝒙𝟐 −𝟒𝒙+𝟒 Evaluate each limit below, (answer will be number, ∞, −∞ or does not exist): 188 𝒙−𝟐 189 −𝟐 𝐥𝐢𝐦 = 𝟐 𝐥𝐢𝐦 (𝒙 − 𝟐𝒙 − 𝟑) 𝟑 = 𝒙→−𝟐 (𝟐+𝒙)𝟐 𝒙→−𝟏− 71 | P a g e 190 𝒙𝟐 +𝟐𝒙−𝟏 191 𝒆𝒙 𝐥𝐢𝐦 = 𝐥𝐢𝐦 = 𝒙→−𝟐 𝒙𝟐 −𝟒 𝒙→𝟏 𝒙𝟐 −𝟏 192 𝟏 193 𝟏 𝐥𝐢𝐦 𝒆 = 𝒙 𝐥𝐢𝐦 𝒆 = 𝒙 𝒙→𝟎− 𝒙→𝟎+ 194 𝟏 195 𝐥𝐢𝐦𝝅 𝒕𝒂𝒏 𝒙 = 𝐥𝐢𝐦 𝒆𝒙 = 𝒙→𝟐 𝒙→𝟎 72 | P a g e 196 𝐥𝐢𝐦 𝒄𝒐𝒕 𝒙 = 197 𝐥𝐢𝐦𝝅 𝒙 𝒔𝒆𝒄𝟐 𝒙 = 𝒙→𝟎 𝒙→ 𝟐 198 𝐥𝐢𝐦+ 𝒍𝒏( 𝒙) = 199 𝐥𝐢𝐦 𝒆𝒔𝒊𝒏 𝒙 = 𝒙→𝟎 𝒙→𝟎 200 𝐥𝐢𝐦+ 𝒕𝒂𝒏−𝟏 (𝒍𝒏( 𝒙)) = 201 𝐥𝐢𝐦 𝒍𝒏( 𝒙 𝒔𝒊𝒏 𝒙) = 𝒙→𝟎 𝒙→𝟎+ 73 | P a g e 202 −𝟐 203 −𝟏 𝐥𝐢𝐦+𝒆 𝟑𝒙 = 𝐥𝐢𝐦 𝒔𝒊𝒏( 𝒆 ) = 𝒙𝟐 𝒙→𝟎 𝒙→𝟎 204 𝐥𝐢𝐦𝝅 𝒆−𝒕𝒂𝒏𝒙 = 205 𝟏 𝐥𝐢𝐦+ 𝒙𝒍𝒏𝒙 = 𝒙→ 𝒙→𝟎 𝟐 Limits at infinity: (horizontal asymptotes( Note: if 𝐥𝐢𝐦 𝒇(𝒙) = 𝑳 , 𝐥𝐢𝐦 𝒇(𝒙) = 𝑳 → 𝒇(𝒙) has horizontal asymptote 𝒙→∞ 𝒙→−∞ 𝒕𝒉𝒆𝒏 𝒚 = 𝑳. Important notes: 1) 𝐥𝐢𝐦 𝑲 = 𝑲 𝒌≠𝟎 𝒙→∓∞ 𝑲 2) 𝐥𝐢𝐦 =𝟎 𝒌≠𝟎 𝒙→∓∞ 𝒙𝒏 ∞ 𝒂>𝟎 3) 𝐥𝐢𝐦 𝒂𝒙𝒏 = { n: positive integer number. 𝒙→∓∞ −∞ 𝒂𝟎 ∞ 𝒂>𝟎 4) 𝐥𝐢𝐦 𝒂𝒙𝒏 = { , 𝐥𝐢𝐦 𝒂𝒙𝒏 = { 𝒙→−∞ ∞ 𝒂