Limit of a function and Limit Laws

2.2: Limit of a function and Limit Laws: The definition of a limit describes what happens to the function values, 𝑓(𝑥) when 𝑥 is near, but not at the value a. It is not affected by whether or how 𝑓(𝑎) is defined. Also, the definition implies that the function values cannot approach two different numbers, so that if a limit exists, it is unique. Note: For a limit to exist, the function must approach the same value from each direction. A Graphical Approach: Use the graph to evaluate the limit. Numerical Algebraic -1=11×1 Graphical µn=3] = * 1¥ . Limit / ✗ 1=1 ✗→ I graph offlx) 4,2, ¢ i. 1 Uni f( . ✗I ✗→ 3 him 2. f- 1×7 1 = = 2 As ✗ LHL = 3- → As ✗ → birth ? y , zt = 4leni 1-1×7 , → I y → 2 ✗→ Uni fix ) 3. ✗→ DHE -3 RHL = yt leni f- 1×7 = 2 ✗→ LHL = 1 = - ✗→ → n . him f- 1×1=1 ✗ → I RHL = leni 1-1×1 ✗→ - 3+ = 2 1LHL YRHL_ A Graphical Approach: Example 2: Determine whether lim ℎ(𝑥) exists, where 𝑥→1 2×1-2 2𝑥 + 2, 𝑥 < 1 ℎ(𝑥) = { . 2𝑥 − 4, 𝑥 ≥ 1 ✗ < I , • ✗ 7, I 2×-4 , Uni 11×7 DHE ✗→ I LHL f- RHL at ✗ LI As 𝑥 → 1− , ℎ(𝑥) → 4 => 𝑥 → 1− ℎ(𝑥) = 2𝑥 + 2 RHI 0 2 . lim ℎ(𝑥) = 4, or 𝐿𝐻𝐿 = 4 0.5 3 2 0 1 𝑥→1− 0.8 3.6 As 𝑥 → 1+ : ℎ(𝑥) → −2 => 𝑥 → 1+ ℎ(𝑥) = 2𝑥 − 4 = 1.8 -0.4 0.9 3.8 0.99 3.98 0.999 3.998 0.99999 3.99998 lim ℎ(𝑥) = −2, or 𝑅𝐻𝐿 = −2 𝑥→1+ 1.1 -1.8 1.01 -1.98 1.001 -1.998 1.0001 -1.9998 Non- existence of a Limit: Example1: None of the following graphs has a limit as 𝑥 approaches 0. Example 2: a) 𝑓(𝑥) = 𝑥 2 −1 𝑥−1 {𝑓(𝑐) is not defined} 𝑥 2 −1 (𝑥 − 1)2 + 2, 𝑥 < 3 b) 𝑓(𝑥) = { −(𝑥 − 3)2 + 1, 𝑥 ≥ 3 ,𝑥 ≠ 1 c) 𝑓(𝑥) = { 𝑥−1 1, 𝑥 = 1 { lim 𝑓(𝑥)does not exist} { lim 𝑓(𝑥) ≠ 𝑓(𝑐)} 𝑥→𝑐 𝑥→𝑐 An Algebraic Approach: 2.2 Limit of a Function, Using Algebraic Simplification Properties of Limits: If L, M, c, and k are real numbers and if lim 𝑓(𝑥) = 𝐿 and 𝑥→𝑐 1. 2. lim 𝑔(𝑥) = 𝑀 , then 𝑥→𝑐 lim(𝑓(𝑥) ± 𝑔(𝑥)) = 𝐿 ± 𝑀 : Sum Rule 𝑥→𝑐 lim (𝑓(𝑥) ∙ 𝑔(𝑥)) = 𝐿 ∙ 𝑀 𝑥→𝑐 3. lim (𝑘 ∙ 𝑓(𝑥)) = 𝑘 ∙ 𝐿 𝑥→𝑐 4. lim 𝑓(𝑥) 𝑥→𝑐 𝑔(𝑥) = 𝐿 𝑀 , 𝑀≠0 : Product Rule : Constant Multiple Rule : Quotient Rule 5. lim (𝑓(𝑥))𝑛 = 𝐿𝑛 , 𝑛 a positive integer : Power Rule 𝑥→𝑐 𝑛 + 6. lim 𝑛√𝑓(𝑥) = √𝐿 = 𝐿1⁄𝑛 , 𝑛 a positive integer: nth Root Rule 𝑥→𝑐 Evaluating Limits of Polynomials and Rational Functions: Theorem 1: For any polynomial, 𝑝(𝑥) = 𝑎𝑛 𝑥 𝑛 + 𝑎𝑛−1 𝑥 𝑛−1 + ⋯ 𝑎0 , lim 𝑝(𝑥) = 𝑝(𝑐). 𝑥→𝑐 Theorem 2: For any polynomials, 𝑝(𝑥) and 𝑞(𝑥), where 𝑞(𝑐) ≠ 0, lim 𝑝(𝑥) 𝑥→𝑐 𝑞(𝑥) = 𝑝(𝑐) 𝑞(𝑐) . Evaluate the following limits. 1. lim (𝑥 3 + 4𝑥 2 − 3) (-213+41-2) 2-3 = 𝑥→−2 - = = 2. lim 𝑥→−1 - 1) 4. 𝑥 2 +5 lim (√4𝑥 2 − 3) 𝑥→0 16 -3 ⑤ = ← 𝑥→−2 lim + 𝑥 3 +4𝑥 2 −2 = 3. 8 = - 𝑥 4 +𝑥 2 −1 𝑥 2 +5 = -1m¥ =/ 513 ✗ 4×-2 𝟎 Indeterminate Forms(𝐚 lim 1. 𝟎 = 'Y 𝑥 2 +𝑥−2 E¥] = a % form = = =¥; ×¥ ¥ " lim 𝑥 2 +𝑥−6 𝑥→2 𝑥 2 −7𝑥+10 ⇐ MY I ÷ = a.CI?-s-) leni ✗→ = - 𝐟𝐨𝐫𝐦): Find the indicated limit if it exits. 𝑥→1 𝑥 2 −𝑥 2. ☒ 2) ( ✗ 1) + ÷? f§] I ¥ = 3=D Rationalize 3. lim 𝑥−4 𝑥→4 √𝑥−2 5¥ ( = a denominator the 8- : ) torn ¥÷ - =¥¥:* : = = ai ✗ →4 ,= J×, = hi ✗→ = u (5×-72) ① 4. lim 5. lim 𝑥→3 𝑥→0 𝑥 4 −81 𝑥−3 2𝑥 2 −3𝑥 𝑥 = leni ✗ → 0 (2,1--3) ✗ = Ani ✗→ = 0 (2×-3) 1-32 lim 6. :¥¥÷¥→ 4−𝑥 2 𝑥→−2 2𝑥 2 +𝑥 3 = Air = ✗ → -2 2¥ = . ÷ = = R lim 7. 𝑥→1 e 1h √𝑥−1 𝑥−1 ( a ¥ ÷ ☐ r: try , 5¥ . = , = = ×¥¥¥=¥¥¥ , hi i¥i¥ = , ✗ →1 8. lim 𝑥→3 12. 𝑓(𝑥) = { fein 𝑥 2 −𝑥−1 √𝑥+1 −2𝑥 + 3, 𝑥 2, 𝑥<1 . Find lim 𝑓(𝑥). 𝑥→1 𝑥≥1 LHL=µñfl× fix ) ) = ✗ → 1- I ✗→ leni i ✗→ (2×+3) = RH thistle = It ✗→ ✗→ = LHL = R HL = leni (F) 1- I i 1¥,f" 2𝑥 − 4, 𝑥 ≤ 3 13. ℎ(𝑥) = { 2 . Find lim ℎ(𝑥). 𝑥→3 𝑥 − 2𝑥, 𝑥 > 3 =D 2-2 The Squeeze Theorem/ The Sandwich Theorem: ' : Note: The graph of 𝑓 is sandwiched between the graphs of 𝑔 and ℎ. Example 10/ Page 52: Given a function 𝑢 that satisfies 1 − 𝑥2 4 ≤ 𝑢(𝑥) ≤ 1 + 𝑥2 2 for all 𝑥 ≠ 0. Find lim 𝑢(𝑥), no matter how complicated 𝑢 is. 𝑥→0 bin =/ UCH ✗ → 0 am f-E) ✗→ 0 am ✗→ 0 (1-1×-2)=1 Note: Any function 𝑢(𝑥) whose graph lies between the graphs of 𝑦 = 1 − 𝑥2 4 and 𝑦 = 1 + limit 1 as 𝑥 → 0. #63/ Page 54: Given a function 𝑓(𝑥) that satisfies √5 − 2𝑥 2 ≤ 𝑓(𝑥) ≤ √5 − 𝑥 2 for all 𝑥: − 1 ≤ 𝑥 ≤ 1. Find lim 𝑓(𝑥). 𝑥→0 Solution: lim 𝑓(𝑥) = √5. = 55 = is 𝑥→0 1%15-7×1 1%4-5*7 I = = 55 𝑥2 2 has