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Ex 13.1, 23 - Find lim x->0 and lim x->1 where f(x) = { 2x + 3, 3(x+1)

Ex 13.1, 23 - Chapter 13 Class 11 Limits and Derivatives - Part 2

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Ex 13.1, 23 - Chapter 13 Class 11 Limits and Derivatives - Part 3 Ex 13.1, 23 - Chapter 13 Class 11 Limits and Derivatives - Part 4

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Ex 13.1, 23 (Method 1) Find lim┬(x→0) f(x) and lim┬(x→1) f(x), where f(x) = {█([email protected](x+1),)┤ ■8(x ≤[email protected]>0) Finding limit at x = 0 lim┬(x→0) f(x) = lim┬(〖x→0〗^− ) f(x) =lim┬(〖x→0〗^+ ) f(x) ∴ (𝒍𝒊𝒎)┬(𝐱→𝟎) f(x) = 3 (𝒍𝒊𝒎)┬(〖𝐱→𝟎〗^− ) f(x) = lim┬(x→0) (2x + 3) = 2(0) + 3 = 0 + 3 = 3 (𝒍𝒊𝒎)┬(〖𝐱→𝟎〗^+ ) f(x) = lim┬(x→0) 3(x + 1) = 3(0 + 1) = 3(1) = 3 Finding limit at x = 1 "f(x) = " {█([email protected](x+1),)┤" " ■8(x ≤[email protected]>0) For x > 0, the function is same Therefore, (𝒍𝒊𝒎)┬(𝒙 → 𝟏) " f(x) " " = " lim┬(x→1) " 3(x + 1) " " = 3(1 + 1)" " = 3 (2)" " = 6" (𝐥𝐢𝐦)┬(𝐱→𝟏) f(x) = 6 Ex 13.1, 23 (Method 2) Find lim┬(x→0) f(x) and lim┬(x→1) f(x), where f(x) = {█([email protected](x+1),)┤ ■8(x ≤[email protected]>0) Finding limit at x = 0 lim┬(x→0) f(x) = lim┬(〖x→0〗^− ) f(x) =lim┬(〖x→0〗^+ ) f(x) LHL at x → 0 lim┬(x→0^− ) f(x) = lim┬(h→0) f(0 − h) = lim┬(h→0) f(−h) = lim┬(h→0) 2(–h) + 3 = 2(0) + 3 = 3 RHL at x → 0 lim┬(x→0^+ ) f(x) = lim┬(h→0) f(0 + h) = lim┬(h→0) f(h) = lim┬(h→0) 3(h + 1) = 3(0 + 1) = 3 ∴ (𝒍𝒊𝒎)┬(𝐱→𝟎) f(x) = 3 Finding limit at x = 1 "f(x) = " {█([email protected](x+1),)┤" " ■8(x ≤[email protected]>0) For x > 0, the function is same Therefore, (𝒍𝒊𝒎)┬(𝒙 → 𝟏) " f(x) " " = " lim┬(x→1) " 3(x + 1) " " = 3(1 + 1)" " = 3 (2)" " = 6" (𝐥𝐢𝐦)┬(𝐱→𝟏) f(x) = 6

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.