CH.1_SALMA.pdf

MATH 101 Calculus I Notation: 𝒙 → 𝒂− (𝑥 approaches 𝑎 from the left) 𝒙 → 𝒂+ (𝑥 approaches 𝑎 from the right) Beware: −𝒂 ≠ 𝒂− +𝒂 ≠ 𝒂+ Only appear in limits The One-sided Limits 𝒍𝒊𝒎 𝒇 𝒙 𝒊𝒔 𝒄𝒂𝒍𝒍𝒆𝒅 𝒍𝒆𝒇𝒕 𝒍𝒊𝒎𝒊𝒕 𝒍𝒊𝒎 𝒇 𝒙 𝒊𝒔 𝒄𝒂𝒍𝒍𝒆𝒅 𝒓𝒊𝒈𝒉𝒕 𝒍𝒊𝒎𝒊𝒕 𝒙→𝒂−...

MATH 101 Calculus I Notation: 𝒙 → 𝒂− (𝑥 approaches 𝑎 from the left) 𝒙 → 𝒂+ (𝑥 approaches 𝑎 from the right) Beware: −𝒂 ≠ 𝒂− +𝒂 ≠ 𝒂+ Only appear in limits The One-sided Limits 𝒍𝒊𝒎 𝒇 𝒙 𝒊𝒔 𝒄𝒂𝒍𝒍𝒆𝒅 𝒍𝒆𝒇𝒕 𝒍𝒊𝒎𝒊𝒕 𝒍𝒊𝒎 𝒇 𝒙 𝒊𝒔 𝒄𝒂𝒍𝒍𝒆𝒅 𝒓𝒊𝒈𝒉𝒕 𝒍𝒊𝒎𝒊𝒕 𝒙→𝒂− 𝒙→𝒂+ Two-sided Limit One-sided Limits 𝑰𝒇 𝐥𝐢𝐦− 𝒇 𝒙 ≠ 𝐥𝐢𝐦+ 𝒇 𝒙 ⟹ 𝐥𝐢𝐦 𝒇 𝒙 = 𝒅𝒐𝒆𝒔𝒏′ 𝒕 𝒆𝒙𝒊𝒔𝒕 𝑫𝑵𝑬 𝒙→𝒂 𝒙→𝒂 𝒙→𝒂  𝑓 𝑎 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑓 𝑎 =2 ☟ 𝑓 𝑎 =1 lim− 𝑓 𝑥 = 1 𝑎𝑛𝑑 lim+ 𝑓 𝑥 = 3 lim 𝑓 𝑥 = 1 𝑎𝑛𝑑 lim+ 𝑓 𝑥 = 3 lim 𝑓 𝑥 = 1 𝑎𝑛𝑑 lim+ 𝑓 𝑥 = 3 𝑥→𝑎 𝑥→𝑎 𝑥→𝑎− 𝑥→𝑎 𝑥→𝑎− 𝑥→𝑎 ⟹One-sided Limits are NOT equal ⟹One-sided Limits are NOT equal ⟹One-sided Limits are NOT equal ⟹ Two-sided Limit doesn’t exist ⟹ Two-sided Limit doesn’t exist ⟹ Two-sided Limit doesn’t exist lim 𝑓 𝑥 = DNE lim 𝑓 𝑥 = DNE lim 𝑓 𝑥 = DNE 𝑥→𝑎 𝑥→𝑎 𝑥→𝑎  ☟ 𝑓 𝑎 =3 𝑓 𝑎 =2 𝑓 𝑎 = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑 lim− 𝑓 𝑥 = 2 = lim+ 𝑓 𝑥 lim 𝑓 𝑥 = 2 = lim+ 𝑓 𝑥 lim− 𝑓 𝑥 = 2 = lim+ 𝑓 𝑥 𝑥→𝑎 𝑥→𝑎 𝑥→𝑎− 𝑥→𝑎 𝑥→𝑎 𝑥→𝑎 ⟹ One-sided Limits are equal ⟹ One-sided Limits are equal ⟹ One-sided Limits are equal ⟹ Two-sided Limit exist ⟹ Two-sided Limit exist ⟹ Two-sided Limit exist lim 𝑓 𝑥 = 2 lim 𝑓 𝑥 = 2 lim 𝑓 𝑥 = 2 𝑥→𝑎 𝑥→𝑎 𝑥→𝑎 Method of working with Limit: 1. Graphical Method 2. Numerical Method 3. Algebraically Solution: Graph: Conjecture: sin 𝑥 sin 𝑥 lim− = 1 = lim+ 𝑥→0 𝑥 𝑥→0 𝑥 sin 𝑥 ⟹ lim =1 𝑥→0 𝑥 Solution: The function 𝑓 𝑥 = 𝑥−1 is undefined at 𝑥 = 1 𝑥−1 𝑥−1 𝑥−1 lim− = 2 = lim+ 𝑥→1 𝑥−1 𝑥→1 𝑥−1 𝑥−1 ⟹ lim =2 𝑥→1 𝑥−1  𝒙 (classic) 𝐅𝐢𝐧𝐝 𝒍𝒊𝒎 𝒙→𝟎 𝒙 𝑥 1 𝑖𝑓 𝑥 > 0 Solution: 𝑓 𝑥 = = ቐ𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑 𝑖𝑓 𝑥 = 0 𝑥 −1 𝑖𝑓 𝑥 < 0 𝑥 𝑥 lim− = −1 𝑎𝑛𝑑 lim+ =1 𝑥→0 𝑥 𝑥→0 𝑥 𝑥 ⟹ lim = 𝐷𝑁𝐸 𝑥→0 𝑥 Infinite Limits: 𝟏 (classic) 𝐅𝐢𝐧𝐝 𝒍𝒊𝒎 𝒙→𝟎 𝒙 𝟏 Solution: 𝑓 𝑥 = 𝒙 1 1 lim−= −∞ 𝑎𝑛𝑑 lim+ =∞ 𝑥→0 𝑥 𝑥→0 𝑥 1 ⟹ lim = 𝐷𝑁𝐸 𝑥→0 𝑥 Infinite Limits: Various cases: −∞ decrease without bound 𝒍𝒊𝒎− 𝒇 𝒙 = ൜ increase without bound 𝒙→𝒂 ∞ 𝒙 → 𝒂+ 𝒙→𝒂 increase without bound decrease without bound increase without bound from the right and left Infinite Limits: Various cases: −∞ decrease without bound 𝒍𝒊𝒎− 𝒇 𝒙 = ൜ increase without bound 𝒙→𝒂 ∞ 𝒙 → 𝒂+ 𝒙→𝒂 decrease without bound increase without bound decrease without bound from the right and left Vertical Asymptotes: of a function 𝑓 Example: Vertical ☟ Asymptote Vertical Vertical ☟ Vertical ☟ Asymptote ☟ Asymptote Asymptote Vertical Asymptotes: Example: 𝒙 = 𝟎 𝒊𝒔 𝒂 Vertical Asymptote of the graph of 𝒇 𝒙 = 𝟏 𝒙 Vertical Asymptote Example: 𝒙 = 𝟏 𝒊𝒔 𝒂 Vertical Asymptote of the graph of 𝒇 𝒙 = 𝟏 𝟐 Vertical Asymptote 𝒙−𝟏 1 1 𝑥−1 2 Example: =𝟑 =𝟐 𝟐𝟎𝟏𝟕 =− 𝟏𝟏 =𝟎 𝝅 = 𝟐 Example: = 𝐥𝐢𝐦 𝒇 𝒙 − 𝒍𝒊𝒎 𝒈 𝒙 + 𝟑 𝒍𝒊𝒎 𝒉 𝒙 𝒙→𝒂 𝒙→𝒂 𝒙→𝒂 𝟐 = 𝟒 − −𝟑 + 𝟑 𝟑 =𝟒+𝟑+𝟐 =𝟗 Example: 𝐥𝐢𝐦 𝒇 𝒙 𝐥𝐢𝐦 𝒉 𝒙 𝒙→𝒂 𝒙→𝒂 = 𝒍𝒊𝒎 𝒈 𝒙 𝒙→𝒂 𝟐 𝟒 = 𝟑 −𝟑 𝟖 = 𝟑 −𝟑 𝟖 = −𝟗 Example: = 𝐥𝐢𝐦 𝒇 𝒙 = 𝟒=𝟐 𝒙→𝒂 Example: = 𝐥𝐢𝐦 𝒙𝟐 − 𝟒 𝒍𝒊𝒎 𝒙 + 𝒍𝒊𝒎 𝟑 𝒙→𝟓 𝒙→𝟓 𝒙→𝟓 = 𝟓𝟐 − 𝟒(𝟓) + 𝟑 = 𝟐𝟓 − 𝟐𝟎 + 𝟑 =𝟖 Example: Solution: 𝐥𝐢𝐦 𝒙𝟕 − 𝟐𝒙𝟓 + 𝟏 𝟑𝟓 = 𝟏𝟕 −𝟐 𝟏 𝟓 +𝟏 𝟑𝟓 𝒙→𝟏 = 𝟐−𝟐 𝟑𝟓 =𝟎 Limit of Rational Functions: Polynomial Polynomial There are 3 cases to consider: Case 1: Case 1: Example: 𝟐𝟐 − 𝟐 + 𝟏 𝟒 − 𝟐 + 𝟏 𝟑 = = = = −𝟑 𝟐−𝟑 −𝟏 −𝟏 (−𝟏)𝟑 −𝟔 −𝟏 − 𝟔 −𝟕 = 𝟐 = = =𝟕 (−𝟏) +𝟐(−𝟏) 𝟏−𝟐 −𝟏 𝟓𝒙𝟑 + 𝟒 𝟓(𝟐)𝟑 +𝟒 𝟒𝟎 + 𝟒 𝟒𝟒 𝒄 𝐥𝐢𝐦 = = = = −𝟒𝟒 𝒙→𝟐 𝒙 − 𝟑 𝟐−𝟑 −𝟏 −𝟏 Case 2: 𝑵𝒐𝒏𝒛𝒆𝒓𝒐 𝒁𝒆𝒓𝒐 Solution: ∞ or −∞ or Doesn’t Exist Example: 𝟐−𝟒 −𝟐 = = = −∞ (𝟒 − 𝟒)(𝟒 + 𝟐) 𝟎+ 𝟐−𝟒 −𝟐 = = − =∞ (𝟒 − 𝟒)(𝟒 + 𝟐) 𝟎 = 𝑫𝑵𝑬 Case 3: 𝟎 𝟎 Solution: 𝑭𝒂𝒄𝒕𝒐𝒓𝒊𝒏𝒈 or rationalizing the numerator or denominator Example: 𝟗 − 𝟏𝟖 + 𝟗 𝟎 = (𝟑 − 𝟑) = 𝟎  𝑨𝟐 − 𝟐𝑨𝑩 + 𝑩𝟐 = 𝑨 − 𝑩 𝟐 (𝒙 − 𝟑)(𝒙 − 𝟑) = 𝐥𝐢𝐦 = 𝐥𝐢𝐦 𝒙 − 𝟑 = 𝟑 − 𝟑 = 𝟎 𝒙→𝟑 (𝒙 − 𝟑) 𝒙→𝟑 Case 3: 𝟎 𝟎 Solution: 𝑭𝒂𝒄𝒕𝒐𝒓𝒊𝒏𝒈 or rationalizing the numerator or denominator Example: −𝟖 + 𝟖 𝟎 (𝟏𝟔 − 𝟒 − 𝟏𝟐) 𝟎  = = (𝟐𝒙 + 𝟖) = 𝐥𝐢𝐦 𝒙→−𝟒 (𝒙 + 𝟒)(𝒙 − 𝟑) 𝟐(𝒙 + 𝟒) 𝟐 𝟐 𝟐 = 𝐥𝐢𝐦 = 𝐥𝐢𝐦 = =− 𝒙→−𝟒 (𝒙 + 𝟒)(𝒙 − 𝟑) 𝒙→−𝟒 (𝒙 − 𝟑) (−𝟒 − 𝟑) 𝟕 Case 3: 𝟎 𝟎 Solution: 𝑭𝒂𝒄𝒕𝒐𝒓𝒊𝒏𝒈 or rationalizing the numerator or denominator Example: 𝟐𝟓 − 𝟏𝟓 − 𝟏𝟎 𝟎 = = 𝟐𝟓 − 𝟓𝟎 + 𝟐𝟓 𝟎  (𝒙 − 𝟓)(𝒙 + 𝟐) (𝒙 + 𝟐) 𝟕 = 𝐥𝐢𝐦 = 𝐥𝐢𝐦 = Case 2 𝒙→𝟓 (𝒙 − 𝟓)(𝒙 − 𝟓) 𝒙→𝟓 (𝒙 − 𝟓) 𝟎 𝒙𝟐 − 𝟑𝒙 − 𝟏𝟎 𝒙+𝟐 𝟕 𝐥𝐢𝐦+ 𝟐 = 𝐥𝐢𝐦+ = + = +∞ 𝒙→𝟓 𝒙 − 𝟏𝟎𝒙 + 𝟐𝟓 𝒙→𝟓 𝒙 − 𝟓 𝟎 𝒙𝟐 − 𝟑𝒙 − 𝟏𝟎 𝐥𝐢𝐦 = 𝑫𝑵𝑬 𝒙𝟐 − 𝟑𝒙 − 𝟏𝟎 𝒙+𝟐 𝟕 𝒙→𝟓 𝒙𝟐 − 𝟏𝟎𝒙 + 𝟐𝟓 𝐥𝐢𝐦− 𝟐 = 𝐥𝐢𝐦− = − = −∞ 𝒙→𝟓 𝒙 − 𝟏𝟎𝒙 + 𝟐𝟓 𝒙→𝟓 𝒙 − 𝟓 𝟎 Recall: Example: Solution: lim 𝒙−𝟏 = 𝟏−𝟏 = 𝟎  𝒙→𝟏 𝒙 − 𝟏 𝟏−𝟏 𝟎 𝒎𝒖𝒍𝒕𝒊𝒑𝒍𝒚 𝒖𝒑 & 𝒅𝒐𝒘𝒏 𝒃𝒚 𝒕𝒉𝒆 𝒄𝒐𝒏𝒋𝒖𝒈𝒂𝒕𝒆 𝒐𝒇 𝒙−𝟏 𝒙−𝟏 𝒙+𝟏 𝒙 − 𝟏 which is 𝒙 + 𝟏 lim = lim 𝒙→𝟏 𝒙 − 𝟏 𝒙→𝟏 𝒙−𝟏 𝒙+𝟏 𝒙−𝟏 𝒙+𝟏 = lim 𝒙→𝟏 𝒙 + 𝒙 − 𝒙 − 𝟏 𝒙−𝟏 𝒙+𝟏 = lim 𝒙→𝟏 𝒙−𝟏 = lim 𝒙+𝟏 = 𝟏+𝟏=𝟐 𝒙→𝟏  Limit of Piecewise-Defined Functions: 1 Example: 𝑥+2 𝑥2 − 5 𝑥 + 13 𝟏 𝟏 𝟏 𝟐 𝐥𝐢𝐦− 𝒇(𝒙) = 𝐥𝐢𝐦−(𝒙𝟐 − 𝟓 ) = 𝟗 − 𝟓 = 𝟒 𝐥𝐢𝐦 𝒇 𝒙 = 𝐥𝐢𝐦− = = − = −∞ 𝐥𝐢𝐦 𝒇 𝒙 = 𝐥𝐢𝐦 (𝒙 − 𝟓) 𝒙→𝟑 𝒙→𝟑 𝒙→−𝟐− 𝒙→−𝟐 𝒙+𝟐 −𝟐 + 𝟐 𝟎 𝒙→𝟎 𝒙→𝟎 𝐥𝐢𝐦+ 𝒇 𝒙 = 𝐥𝐢𝐦+(𝒙𝟐 − 𝟓) = (−𝟐)𝟐 −𝟓 = 𝟒 − 𝟓 = −𝟏 =𝟎−𝟓 𝐥𝐢𝐦 𝒇(𝒙) = 𝐥𝐢𝐦+ 𝒙 + 𝟏𝟑 = 𝟏𝟔 = 𝟒 𝒙→𝟑+ 𝒙→𝟑 𝒙→−𝟐 𝒙→−𝟐 = −𝟓 𝐥𝐢𝐦 𝒇(𝒙) = 𝑫𝑵𝑬 𝐥𝐢𝐦 𝒇(𝒙) = 𝟒 𝒙→−𝟐 𝒙→𝟑  𝑛 𝑛 lim 𝑓 𝑥 (e) 𝑥→∞ = lim 𝑓 𝑥 𝑥→∞ 𝑥→∞ 𝑥→∞ true The End Behavior of 𝒇 𝒙 as 𝒙 → ±∞ Various cases: 𝒍𝒊𝒎 𝒇 𝒙 = ∞ 𝒍𝒊𝒎 𝒇 𝒙 = ∞ 𝒙→−∞ 𝒙→∞ 𝒍𝒊𝒎 𝒇 𝒙 = −∞ 𝒍𝒊𝒎 𝒇 𝒙 = −∞ 𝒙→−∞ 𝒙→∞ Basic limits as 𝒙 → ±∞ ➀ 𝒍𝒊𝒎 𝒌 = 𝒌 𝒙→±∞ 𝟏 ② 𝒍𝒊𝒎 =𝟎 Note: (infinite limits) 𝒙→±∞ 𝒙 𝒌  ∞=𝟎 𝟏 𝒌 ➂ If 𝒂 > 𝟎, 𝒕𝒉𝒆𝒏 𝒍𝒊𝒎 =𝟎 𝒂𝒏𝒅 = 𝟎, for 𝒌 ∈ ℝ and 𝒂 ∈ ℤ+ 𝒍𝒊𝒎  𝒌 =∞ 𝒙→±∞ 𝒙𝒂 𝒙→±∞ 𝒙 𝒂 𝟎 𝒍𝒊𝒎 𝒙 = ∞ ➃ 𝒙→∞ 𝒂𝒏𝒅 𝒍𝒊𝒎 𝒙 = −∞ 𝒙→−∞  𝒌+∞=∞ 𝒍𝒊𝒎 𝒙𝒏 = ∞, 𝒏 = 𝟏, 𝟐, 𝟑, 𝟒, ⋯  𝒌 − ∞ = −∞ ➄ 𝒙→∞ −∞, 𝑛 = 1,3,5, ⋯ (ODD) ➅ 𝒍𝒊𝒎 𝒙𝒏 = ൜ 𝒙→−∞ ∞, 𝑛 = 2, 4, 6, ⋯ (EVEN)  Basic limits as 𝒙 → ±∞ Example: ➄ ➅ ➄ ➅ (a) 𝒍𝒊𝒎 𝒙𝟐 = ∞ (b) 𝒍𝒊𝒎 𝒙𝟐 = ∞ (c) 𝒍𝒊𝒎 𝒙𝟑 = ∞ (d) 𝒍𝒊𝒎 𝒙𝟑 = −∞ 𝒙→∞ 𝒙→−∞ 𝒙→∞ 𝒙→−∞ ➅ (e) 𝒍𝒊𝒎 𝟐𝒙𝟓 = 𝟐 𝒍𝒊𝒎 𝒙𝟓 (f) 𝒍𝒊𝒎 𝟐𝒙𝟓 = 𝟐 𝒍𝒊𝒎 𝒙𝟓 𝒍𝒊𝒎 −𝟕𝒙𝟔 = −𝟕 ⋅ ∞ (g) 𝒍𝒊𝒎 −𝟕𝒙𝟔 = −𝟕 𝒍𝒊𝒎 𝒙𝟔 (h) 𝒙→−∞ 𝒙→∞ 𝒙→∞ 𝒙→−∞ 𝒙→−∞ 𝒙→∞ 𝒙→∞ ➄ ➄ = −∞ =𝟐⋅∞ ➅ = −𝟕 ⋅ ∞ = 𝟐 ⋅ −∞ = −∞ = −∞ =∞ Basic limits as 𝒙 → ±∞ Example: ☟ ☝ 𝝅 𝝅 𝒍𝒊𝒎 𝐭𝐚𝐧−𝟏 𝒙 =− 𝒍𝒊𝒎 𝐭𝐚𝐧−𝟏 𝒙 = 𝒙→−∞ 𝟐 𝒙→∞ 𝟐 Limits of Polynomials as 𝒙 → ±∞ Example: = 𝒍𝒊𝒎 −𝟒𝒙𝟓 = −𝟒 ⋅ ∞ = −∞ 𝒙→∞ = 𝒍𝒊𝒎 −𝟒𝒙𝟓 = −𝟒(−∞) = ∞ 𝒙→−∞ (c) 𝐥𝐢𝐦 (𝟕𝒙𝟓 − 𝟒𝒙𝟑 + 𝟐𝒙 − 𝟗) = 𝒍𝒊𝒎 𝟕𝒙𝟓 = 𝟕(−∞) = −∞ 𝒙→−∞ 𝒙→−∞ Limits of Rational Functions as 𝒙 → ±∞ Method 1: Just find the highest degree in the denominator and divide every term by it. Example: 𝟑𝒙 𝟓 𝟓 𝟓 + 𝟑+ 𝟑 + 𝟑+𝟎 𝟏 = 𝒍𝒊𝒎 𝒙 𝒙 = 𝒍𝒊𝒎 𝒙 = ∞ = = 𝒙→∞ 𝟔𝒙 𝟖 𝒙→∞ 𝟖 𝟖 𝟔 − 𝟎 𝟐 − 𝟔− 𝟔− 𝒙 𝒙 𝒙 ∞ Method 2: Just take the ratio of leading terms. Example: 𝟑𝒙 𝟑 𝟏 = 𝒍𝒊𝒎 = 𝒍𝒊𝒎 = 𝒙→∞ 𝟔𝒙 𝒙→∞ 𝟔 𝟐 Limits of Rational Functions as 𝒙 → ±∞ Example: 𝟓𝒙𝟑 𝟓𝒙𝟐 𝟓 ∞ = 𝒍𝒊𝒎 = 𝒍𝒊𝒎 =− = −∞ 𝒙→−∞ −𝟑𝒙 𝒙→−∞ −𝟑 𝟑 𝟒𝒙𝟐 𝟐 𝟐 = 𝒍𝒊𝒎 = 𝒍𝒊𝒎 = =𝟎 𝒙→−∞ 𝟐𝒙𝟑 𝒙→−∞ 𝒙 −∞ Limits Involving Radicals as 𝒙 → ±∞ Example: 𝑥2 + 2 𝑥2 + 2 = lim 𝑥 = lim 𝑥 𝑥→∞ 3𝑥 − 6 𝑥→∞ 3𝑥 6 − 𝑥 𝑥 𝑥 𝑥2 2 + 𝑥2 𝑥2 = lim 𝑥→∞ 3𝑥 6 − 𝑥 𝑥 2 𝑥 2 = 𝑥 = 𝑥 (since 𝑥 → ∞) + 1+ 2 𝑥 = lim 𝑥→∞ 6 3− 𝑥 2 1+ 𝟎 1 1 ∞ = = = 6 3 3 3−∞𝟎 Limits Involving Radicals as 𝒙 → ±∞ Example: 𝑥2 + 2 𝑥2 + 2 = lim 𝑥 = lim 𝑥 𝑥→−∞ 3𝑥 − 6 𝑥→−∞ 3𝑥 6 − 𝑥 𝑥 𝑥 𝑥2 2 + 𝑥2 𝑥2 = lim 𝑥→−∞ 3𝑥 6 𝑥 −𝑥 2 𝑥 2 = 𝑥 = −𝑥 (since 𝑥 → −∞) − 1+ 2 𝑥 = lim 𝑥→−∞ 6 3− 𝑥 2 − 1 + ∞ 𝟎 − 1 −1 = = = 6 3 3 3− 𝟎 ∞ Limits Involving Radicals as 𝒙 → ±∞ Example: 𝑥 3 − 4𝑥 + 5 + 10 𝑥 3 − 4𝑥 + 5 10 + 2 = lim 𝑥2 = lim 𝑥 2 𝑥 𝑥→∞ 2𝑥 2 − 6 𝑥→∞ 2𝑥 2 6 − 2 𝑥2 𝑥2 𝑥 𝑥 3 4𝑥 5 10 − + + 𝑥4 𝑥4 𝑥4 𝑥2 = lim 𝑥→∞ 2𝑥 2 6 − 2 𝑥2 𝑥 1 4 5 10 − 3+ 4+ 2 𝑥 𝑥 𝑥 𝑥 = lim 𝑥→∞ 6 2− 2 𝑥 1 4 5 10 ∞−∞+∞+ ∞ 0 = = =0 6 2 2− ∞ Review of exponential and logarithmic functions (see page 52-60) Infinite Limits laws for exponential and logarithmic functions:  𝒍𝒊𝒎𝒆𝒙 = ∞ 𝒙→∞  𝒍𝒊𝒎 𝒆𝒙 = 𝟎 𝒙→−∞  𝒍𝒊𝒎𝒆−𝒙 = 𝟎 𝒙→∞  𝒍𝒊𝒎 𝒆−𝒙 = ∞ 𝒙→−∞  𝒍𝒊𝒎𝒍𝒏𝒙 = ∞ 𝒙→∞  𝒍𝒊𝒎+ 𝒍𝒏𝒙 = −∞ 𝒙→𝟎 Horizontal Asymptotes: A horizontal line 𝒚 = 𝑳 is called a horizontal asymptote of the graph of a function 𝒇 if either 𝒍𝒊𝒎 𝒇 𝒙 = 𝑳 or 𝒍𝒊𝒎 𝒇 𝒙 = 𝑳 𝒙→∞ 𝒙→−∞ Example: 𝟏 𝟏 𝒍𝒊𝒎 = = 𝟎 𝒙→∞ 𝒙 ∞ ⟹ 𝒚 = 𝟎 (X-axis) is horizontal asymptote 𝒍𝒊𝒎(𝑥 2 + 1) = ∞ 𝒙→∞ ⟹ There is NO horizontal asymptote 𝟑𝒙 + 𝟕 𝟑𝒙 𝟑 𝟑 𝒍𝒊𝒎 = 𝒍𝒊𝒎 = 𝒍𝒊𝒎 = 𝒙→∞ 𝟓𝒙 − 𝟐 𝒙→∞ 𝟓𝒙 𝒙→∞ 𝟓 𝟓 𝟑 ⟹ 𝒚 = 𝟓 is horizontal asymptote Example: horizontal asymptote ☟ horizontal asymptote 𝝅 𝝅 𝒍𝒊𝒎 𝐭𝐚𝐧−𝟏 𝒙 =− 𝒍𝒊𝒎 𝐭𝐚𝐧−𝟏 𝒙 = 𝒙→−∞ 𝟐 𝒙→∞ 𝟐 𝝅 ⟹ 𝒚 = ± 𝟐 are the horizontal asymptotes  (𝒍𝒊𝒎+ 𝒇 𝒙 = 𝒍𝒊𝒎− 𝒇 𝒙 ) 𝒙→𝒄 𝒙→𝒄 [That is, a function 𝒇 𝒙 is continuous at 𝒙 = 𝒄 if 𝒍𝒊𝒎+ 𝒇 𝒙 = 𝒍𝒊𝒎− 𝒇 𝒙 = 𝒇 𝒄 ] 𝒙→𝒄 𝒙→𝒄 Remark: Types of Discontinuities: removable discon. jump discon. (a) 𝒇 𝒄 𝒊𝒔 𝒏𝒐𝒕 𝒅𝒆𝒇𝒊𝒏𝒆𝒅 (b) 𝒍𝒊𝒎 𝒇 𝒙 ≠ 𝒍𝒊𝒎− 𝒇 𝒙 𝒙→𝒄− 𝒙→𝒄 ⟹ 𝒍𝒊𝒎 𝒇 𝒙 = 𝑫𝑵𝑬 𝒙→𝒄 infinite discon. removable discon. (c) 𝒍𝒊𝒎 𝒇 𝒙 = ∞ (𝑫𝑵𝑬) (d) 𝒍𝒊𝒎 𝒇 𝒙 ≠ 𝒇 𝒄 𝒙→𝒄 𝒙→𝒄 Example: 𝟐𝟐 − 𝟒 𝟎 ① 𝒇 𝟐 = = ① 𝒈 𝟐 =𝟑 𝟐−𝟐 𝟎 ⟹ 𝒇 𝟐 is undefined 𝒙𝟐 − 𝟒 𝟒 − 𝟒 𝟎 𝟐−𝟐 𝟎 ② 𝒍𝒊𝒎 = = ⟹ 𝒇 𝒙 is not continuous at 𝒙 = 𝟐 𝒙→𝟐 𝒙 − 𝟐 (𝒙 + 𝟐)(𝒙 − 𝟐) ⟹ 𝒍𝒊𝒎 = 𝒍𝒊𝒎 𝒙 + 𝟐 𝒙→𝟐 (𝒙 − 𝟐) 𝒙→𝟐 =𝟐+𝟐=𝟒 ➂ 𝒍𝒊𝒎 𝒙→𝟐 𝒈(𝒙) ≠ 𝒈 𝟐 ⟹ 𝒈 𝒙 is not continuous at 𝒙 = 𝟐 Example: ① 𝒉 𝟐 =𝟒 𝒙𝟐 − 𝟒 𝟒 − 𝟒 𝟎 𝟐−𝟐 𝟎 ② 𝒍𝒊𝒎 = = 𝒙→𝟐 𝒙 − 𝟐 (𝒙 + 𝟐)(𝒙 − 𝟐) ⟹ 𝒍𝒊𝒎 = 𝒍𝒊𝒎 𝒙 + 𝟐 = 𝟐 + 𝟐 = 𝟒 𝒙→𝟐 (𝒙 − 𝟐) 𝒙→𝟐 ➂ 𝒍𝒊𝒎 𝒙→𝟐 𝒉 𝒙 =𝒉 𝟐 =𝟒 ⟹ 𝒉 𝒙 is continuous at 𝒙 = 𝟐 Example: 𝟏−𝒙 𝒙𝟐 ① 𝒇 𝟏 = 𝟏𝟐 = 𝟏 𝒍𝒊𝒎− 𝒇(𝒙) = 𝒍𝒊𝒎− (𝟏 − 𝒙) = 𝟏 − 𝟏 = 𝟎 ② 𝒙→𝟏 𝒙→𝟏 𝒍𝒊𝒎 𝒇(𝒙) = 𝑫𝑵𝑬 𝒙→𝟏 𝒍𝒊𝒎+ 𝒇(𝒙) = 𝒍𝒊𝒎+ 𝒙𝟐 = 𝟏 𝒙→𝟏 𝒙→𝟏 ⟹ 𝒇 𝒙 is not continuous at 𝒙 = 𝟏 Example: 𝒙𝟐 + 𝟏 𝟑𝒙 − 𝟒 Solution: ① 𝒇 𝟎 = 𝟑 𝟎 − 𝟒 = −𝟒 ② 𝒍𝒊𝒎 𝒇 𝒙 = 𝑫𝑵𝑬 because 𝒙→𝟎 𝒍𝒊𝒎 𝒇(𝒙) = 𝒍𝒊𝒎+ (𝟑𝒙 − 𝟒) = −𝟒 𝒙→𝟎+ 𝒙→𝟎 While, 𝒍𝒊𝒎− 𝒇(𝒙) = 𝒍𝒊𝒎− (𝒙𝟐 + 𝟏) = 𝟏 𝒙→𝟎 𝒙→𝟎 ⟹ 𝒇 𝒙 is not continuous at 𝒙 = 𝟎 Functions Continuous on an Interval: A function 𝒇 is continuous on 𝒂, 𝒃 or −∞, ∞ if 𝒇 is continuous at each 𝒙 = 𝒄 in the interval. 𝒂 𝑐 𝒃 𝑐 If 𝒇 is continuous on −∞, ∞ , we say that 𝒇 is continuous everywhere. Definition: At 𝒙 = 𝒄, 𝒇 is continuous from the left if 𝒍𝒊𝒎− 𝒇(𝒙) = 𝒇(𝒄) 𝒙→𝒄 𝒇 is continuous from the right if 𝒍𝒊𝒎+ 𝒇(𝒙) = 𝒇(𝒄) 𝒙→𝒄 Functions Continuous on an Interval: Example: Show that 𝒇 𝒙 = 𝟗 − 𝒙𝟐 is continuous on its domain. Sketch the graph of 𝒇 𝒙. Solution: Find the domain: 𝟗 − 𝒙𝟐 ≥ 𝟎 ⟹ 𝟗 ≥ 𝒙𝟐 ⟹ 𝟗 ≥ 𝒙𝟐 ⟹3≥ 𝑥 ⟹ −𝟑 ≤ 𝒙 ≤ 𝟑 ⟹ Domain = −𝟑,𝟑 We shall show that 𝒇 is continuous on −𝟑,𝟑. 𝒇 𝟎 =𝟑 ① At 𝒙 = 𝒄, − 𝟑 < 𝒄 < 𝟑 𝒍𝒊𝒎 𝒇 𝒙 = 𝒍𝒊𝒎 𝟗 − 𝒙𝟐 = 𝟗 − 𝒄𝟐 = 𝒇 𝒄 𝒙→𝒄 𝒙→𝒄 ⟹ 𝒇 continuous on −3,3 ② 𝒍𝒊𝒎+ 𝒇(𝒙) = 𝒍𝒊𝒎+ 𝟗 − 𝒙𝟐 = 𝟗 − 𝟗 = 𝟎 = 𝒇 −𝟑 𝒙→−𝟑 𝒙→−𝟑 ➂ 𝒍𝒊𝒎− 𝒇(𝒙) = 𝒍𝒊𝒎− 𝟗 − 𝒙𝟐 = 𝟗 − 𝟗 = 𝟎 = 𝒇 𝟑 𝒙→𝟑 𝒙→𝟑 Thus, 𝒇 is continuous on its domain −𝟑,𝟑 Example: Solution: We solve 𝒙𝟐 − 𝟓𝒙 + 𝟔 = 𝟎 ⟹ 𝒙−𝟐 𝒙−𝟑 =𝟎 ⟹ 𝒙 = 𝟐 𝒐𝒓 𝒙 = 𝟑 ∴ 𝒚 is discontinuous at 𝒙 = 𝟐 𝒂𝒏𝒅 𝟑 Example: Solution: 𝒙 𝒊𝒇 𝒙 > 𝟎 𝒇 𝒙 = 𝒙 =ቐ 𝟎 𝒊𝒇 𝒙 = 𝟎 −𝒙 𝒊𝒇 𝒙 < 𝟎 Since both 𝒙 and −𝒙 are polynomails, we know 𝒇 is continuous everywhere, except possibly at 𝒙 = 𝟎. We check the continuity at 𝒙 = 𝟎: ① 𝒇 𝟎 = 𝟎 =𝟎 𝒍𝒊𝒎− 𝒇(𝒙) = 𝒍𝒊𝒎− (−𝒙) = 𝟎 ②𝒙→𝟎 𝒙→𝟎 So, 𝒇 𝒙 = 𝒙 is continuous at 𝒙 = 𝟎 𝒍𝒊𝒎 𝒇(𝒙) = 𝒍𝒊𝒎+ (𝒙) = 𝟎 𝒙→𝟎+ 𝒙→𝟎 ➂ 𝒍𝒊𝒎 𝒇 𝒙 = 𝒇 𝟎 = 𝟎 𝒙→𝟎 Theorem: “The composition of continuous functions is also continuous” If 𝒈 is continuous at 𝒙 and 𝒇 is continuous at 𝒈(𝒙), then 𝒇 ∘ 𝒈 is continuous at 𝒙. 𝒈 𝒇 𝑥 𝒈(𝒙) 𝒇(𝒈 𝒙 ) 𝒇∘𝒈 Remark: The absolute value of a continuous function is also continuous. Example: If 𝒇 𝒙 = 𝒙 and 𝒈 𝒙 = 𝟒 − 𝒙𝟐 , show that 𝒇 ∘ 𝒈 is continuous everywhere. Solution: 𝒇∘𝒈 𝒙 =𝒇 𝒈 𝒙 = 𝒇 𝟒 − 𝒙𝟐 = 𝟒 − 𝒙𝟐 Polynomial (continuous everywhere) Then 𝒇 ∘ 𝒈 is continuous everywhere by using the above remark. Example: Solution: 𝒙𝟐 + 𝒌𝒙 = 𝟑 𝒙 𝟑 When 𝒙 = 𝟏 ⟹ 𝟏 𝟐+𝒌 𝟏 = 𝟏 ⟹𝟏+𝒌=𝟑 ⟹𝒌=𝟑−𝟏 ⟹𝒌=𝟐 Example: 𝟏−𝟏 𝟎 = 𝒄𝒐𝒔 = 𝒄𝒐𝒔  𝟏−𝟏 𝟎 (𝒙 − 𝟏)(𝒙 + 𝟏) = 𝒍𝒊𝒎 𝒄𝒐𝒔 = 𝒍𝒊𝒎 𝒄𝒐𝒔 𝒙 + 𝟏 = 𝒄𝒐𝒔 𝟐 𝒙→𝟏 (𝒙 − 𝟏) 𝒙→𝟏 𝟎 = 𝟎 sin 𝟎 + cos 𝟎 = cos 𝟎 = 𝟏 𝟎 Theorem: “The Squeezing Theorem” If 𝒈 𝒙 ≤ 𝒇 𝒙 ≤ 𝒉 𝒙 for all 𝒙 ≠ 𝒄, in some open interval contining 𝒄, and 𝒍𝒊𝒎 𝒈 𝒙 = 𝑳 = 𝒍𝒊𝒎 𝒉 𝒙. 𝒙→𝒄 𝒙→𝒄 Then 𝒍𝒊𝒎 𝒇 𝒙 = 𝑳 too. 𝒙→𝒄 Example: 𝟏 𝟐 ⋅ sin 𝟐𝜽 sin 𝟐𝜽 = lim = 𝟐 lim =𝟐 𝟏 =𝟐 𝜽→𝟎 𝟐⋅𝜽 𝜽→𝟎 𝟐𝜽 𝒔𝒊𝒏 𝒙 𝒔𝒊𝒏 𝒙 𝟏 = 𝒍𝒊𝒎 = 𝒍𝒊𝒎 ⋅ 𝒙→𝟎 𝒙 ⋅ 𝒄𝒐𝒔 𝒙 𝒙→𝟎 𝒙 𝒄𝒐𝒔 𝒙 𝟏 𝒔𝒊𝒏 𝒙 𝟏 = 𝒍𝒊𝒎 𝒍𝒊𝒎 𝒙→𝟎 𝒙 𝒙→𝟎 𝒄𝒐𝒔 𝒙 sec 𝑥 = 𝟏 𝒍𝒊𝒎 𝒔𝒆𝒄 𝒙 𝒙→𝟎 = 𝒔𝒆𝒄 𝟎 =𝟏 Example: 𝟏 sin 𝟑𝒙 sin 𝟑𝒙 ⋅ = lim 𝒙 = lim 𝒙 𝑥→𝟎 𝟏 𝑥→𝟎 sin 𝟓𝒙 sin 𝟓𝒙 ⋅ 𝒙 𝒙 𝟑 sin 𝟑𝒙 = lim 𝟑𝒙 𝑥→𝟎 𝟓 sin 𝟓𝒙 𝟓𝒙 𝟑 sin 𝟑𝒙 lim 𝟑𝒙 = 𝑥→𝟎 𝟓 sin 𝟓𝒙 lim 𝑥→𝟎 𝟓𝒙 𝟏 𝟑 lim sin 𝟑𝒙 = 𝑥→𝟎 𝟑𝒙 𝟓 lim sin 𝟓𝒙 𝑥→𝟎 𝟓𝒙 𝟏 𝟑 = 𝟓 Example: Using squeezing thm. 𝟏 −𝟏 ≤ 𝐬𝐢𝐧 ≤ 𝟏 for all 𝒙 ≠ 𝟎 𝒙 𝟏 −𝟏 ⋅ 𝒙 ≤ 𝒙 ⋅ 𝐬𝐢𝐧 ≤𝟏⋅𝒙 𝒙 1 𝟏 𝒍𝒊𝒎 sin = 𝑫𝑵𝑬 −𝒙 ≤ 𝒙 𝐬𝐢𝐧 ≤𝒙 𝒙→𝟎 𝑥 𝒙 𝟏 As 𝒙 → 𝟎, 𝒔𝒊𝒏 𝟏 oscillates between −𝟏 and 𝟏 𝐥𝐢𝐦 −𝒙 ≤ 𝐥𝐢𝐦 𝒙 𝒔𝒊𝒏 ≤ 𝐥𝐢𝐦 𝒙 𝒙 𝒙→𝟎 𝒙→𝟎 𝒙 𝒙→𝟎 faster and faster. 𝟏 𝟎 ≤ 𝒍𝒊𝒎 𝒙 𝒔𝒊𝒏 ≤𝟎 𝒙→𝟎 𝒙 𝟏 ⟹ 𝒍𝒊𝒎 𝒙 𝒔𝒊𝒏 =𝟎 𝒙→𝟎 𝒙  𝑏>1 0